FS
O
PR
O
PA
G
E
8
EC
TE
D
Integral calculus
8.1 Kick off with CAS
R
8.2 Areas under and between curves
U
N
C
O
R
8.3 Linear substitutions
c08IntegralCalculus.indd 378
8.4 Other substitutions
8.5 Integrals of powers of trigonometric functions
8.6 Integrals involving inverse trigonometric functions
8.7 Integrals involving partial fractions
8.8 Review
14/08/15 4:31 PM
8.1 Kick off with CAS
Exploring integration with CAS
In this topic, we will be integrating various types of functions.
1 Use CAS to find each of the following.
a 3 (3x + 4) 5dx
b 3
c 3
1
dx
(3x + 4) 2
1
"3x + 4
Can you predict 3
1
"a2
− b2x2
1
O
"9 − 16x
2
1
dx
4 + 9x2
a2
b 3
1
dx
9 + 16x2
EC
4 Use CAS to find each of the following.
a 3
1
dx
4 − 9x2
a2
"16 − 25x
2
dx
c 3
1
dx
16 + 25x2
b 3
1
dx
9 − 16x2
c 3
1
dx
16 − 25x2
1
dx, where a and b are positive constants?
− b2x2
U
N
C
O
R
R
Can you predict 3
1
1
dx, where a and b are positive constants?
+ b2x2
TE
D
Can you predict 3
dx
dx, where a and b are positive constants?
3 Use CAS to find each of the following.
a 3
c 3
PR
O
"4 − 9x
2
b 3
dx
E
1
PA
G
a 3
1
FS
Can you predict 3 (ax + b) ndx? What happens when n = –1?
2 Use CAS to find each of the following.
dx
d 3
3x + 4
dx
Please refer to
the Resources
tab in the Prelims
section of your
eBookPLUS for
a comprehensive
step-by-step
guide on how to
use your CAS
technology.
c08IntegralCalculus.indd 379
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8.2
Areas under and between curves
Area bounded by a curve and the x-axis
Basic integration techniques and evaluating areas bounded by curves and the x-axis,
have been covered in the Mathematical Methods course. The examples and theory
presented here are a review of this material.
y
y = f(x)
b
Recall that the definite integral, A = 3f(x)dx, gives a
a
a
b
x
FS
0
O
measure of the area A bounded by the curve y = f(x),
the x-axis and the lines x = a and x = b. This result is
known as the Fundamental Theorem of Calculus.
Find the area bounded by the line y = 2x + 3, the x-axis and the lines x = 2
and x = 6.
THINK
WRITE/DRAW
y
18
16
14
12
10
8
6
4
2
1 Draw a diagram to identify the required
R
EC
TE
D
PA
G
area and shade this area.
O
definite integral.
R
2 The required area is given by a
C
3 Perform the integration using square
U
N
bracket notation.
4 Evaluate.
–6 –4 –2 0
–2
–4
–6
y = 2x + 3
2 4 6 8 10 12 x
6
A = 3 (2x + 3)dx
2
A = c x2 + 3x d
6
2
A = (62 + 3 × 6) − (22 + 3 × 2)
= (36 + 18) − (4 + 6)
= 44
5 State the value of the required area region.
The area is 44 square units.
6 The shaded area is a trapezium. As a
The width of the trapezium is h = 6 − 2 = 4,
and since y = 2x + 3:
when x = 2, y = 7 and when x = 6, y = 15.
check on the result, find the area using the
formula for a trapezium.
380
PR
O
1
E
WORKED
EXAMPLE
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 380
14/08/15 6:14 PM
y
18
16
14
12
10
8
6
4 7
2
15
2 4 6 8 10 12 x
4
FS
–6 –4 –2 0
–2
–4
–6
y = 2x + 3
PR
O
O
h
The area of a trapezium is (a + b).
2
The area is 42 (7 + 15) = 44 square units
WORKED
EXAMPLE
2
PA
G
E
Using symmetry
Sometimes symmetry can be used to simplify the area calculation.
Find the area bounded by the curve y = 16 − x2, the x-axis and the lines
x = ±3.
WRITE/DRAW
TE
D
THINK
1 Factorise the quadratic to find the
EC
x-intercepts.
y = 16 − x2
y = (4 − x)(4 + x)
The graph crosses the x-axis at x = ±4 and
crosses the y-axis at y = 16.
y
18
16
14
12
10
8
6
4
2
2 Draw a diagram to identify the required
U
N
C
O
R
R
area and shade this area.
–6 –5 –4 –3 –2 –1 0
–2
–4
y = 16 – x2
1 2 3 4 5 6 x
3
3 The required area is given by a definite
integral; however, we can use symmetry.
A = 3 (16 − x2)dx
−3
0
3
−3
0
= 3 (16 − x2)dx + 3 (16 − x2)dx
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 381
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14/08/15 6:13 PM
0
3
−3
0
However, 3 (16 − x2)dx = 3 (16 − x2)dx
3
A = 23 (16 − x2)dx
0
3
0
A = 2 c a 16 × 3 − 13 (3) 3 b − 0 d
5 Evaluate.
A = 2(48 − 9)
The area is 78 square units.
PR
O
6 State the value of the required area.
FS
bracket notation.
A = 2 c 16x − 13x3 d
O
4 Perform the integration using square
3
Find the area under one arch of the sine curve y = 5 sin(3x).
TE
D
WORKED
EXAMPLE
PA
G
E
Areas involving basic trigonometric functions
For integrals and area calculations involving the basic trigonometric functions,
1
1
we use the results 3cos(kx)dx = sin(kx) + c and 3sin(kx)dx = − cos(kx) + c,
k
k
where x ∈ R, k ≠ 0, and k and c are constants.
THINK
WRITE/DRAW
1 Draw a diagram to identify the required
U
N
C
O
R
R
EC
area and shade this area.
2 One arch is defined to be the area under
one half-cycle of the sine wave.
y = 5 sin(3x) has an amplitude of 5 and a period
2π
of .
3
y
6
5
4
3
2
1
y = 5 sin(3x)
0
–1
–2
–3
–4
–5
–6
π
–
3
2π
––
3
π x
The graph crosses the x-axis at sin(3x) = 0,
π
2π
when x = 0, and . The required area is
3
3
π
3
A = 35 sin (3x) dx
0
382
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 382
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π
A=
3 Perform the integration using square
bracket notation.
4 Evaluate, taking the constant factors outside
the brackets.
3
5 c −13 cos (3x) d
0
A = −53[cos(π) − cos(0) ]
= −53[−1 − 1]
The area is
5 State the value of the required area.
10
3
square units.
4
Find the area bounded by the coordinate axes, the graph of y = 4e −2x and
the line x = 1.
E
WORKED
EXAMPLE
PR
O
O
FS
Areas involving basic exponential functions
For integrals and area calculations involving the basic exponential functions, we use
1
the result 3ekxdx = ekx + c, where x ∈ R, k ≠ 0, and k and c are constants.
k
WRITE/DRAW
1 Draw a diagram to identify the required
area and shade this area.
PA
G
THINK
y
5
4
3
TE
D
EC
R
R
O
2 The required area is given by a
C
definite integral.
U
N
3 Perform the integration using square
bracket notation.
y = 4e–2x
2
1
0
–1
1
2
3 x
–1
1
A = 34e−2xdx
0
A=
4 c −12e−2x d
1
0
1
= −2 c e−2x d 0
4 Evaluate.
A = −2[ e−2 − e0]
= −2(e−2 − 1)
5 State the value of the required area.
The exact area is 2(1 − e−2) square units.
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 383
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Areas involving signed areas
When evaluating a definite integral, the result is a number; this number can be
positive or negative. A definite integral which represents an area is a signed area; that
is, it may also be positive or negative. However, areas cannot be negative.
Areas above the x-axis
When a function is such that f(x) ≥ 0 for a ≤ x ≤ b, where
b > a, that is, the function lies above the x-axis, then the
definite integral that represents the area A is positive:
y
y = f(x)
b
0
b
PA
G
a
a
b
x
0
E
3f(x)dx < 0.
x
y
PR
O
Areas below the x-axis
When a function is such that f(x) ≤ 0 for a ≤ x ≤ b, where
b > a, that is, the function lies below the x-axis, then the
definite integral that represents the area A is negative:
b
O
a
a
FS
A = 3f(x)dx > 0.
y = f(x)
So, when an area is determined that is bounded by a curve that is entirely below the
x-axis, the result will be a negative number. Because areas cannot be negative, the
absolute value of the integral must be used.
b
b
a
a
b
TE
D
A = 3 3f(x)dx 3 = −3f(x)dx = 3f(x)dx
a
5
Find the area bounded by the curve y = x2 − 4x + 3 and the x-axis.
EC
WORKED
EXAMPLE
R
THINK
R
1 Factorise the quadratic to
C
O
find the x-intercepts.
WRITE/DRAW
y = x2 − 4x + 3
y = (x − 3)(x − 1)
The graph crosses the x-axis at x = 1 and x = 3 and
crosses the y-axis at y = 3.
y
5
U
N
2 Sketch the graph, shading
the required area.
4
3
2
1
–1
0
1
2
3
4
5 x
–1
384
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 384
14/08/15 6:14 PM
3
A = 3 3 (x2 − 4x + 3)dx 3
3 The required area is below
the x-axis and will evaluate
to a negative number. The
area must be given by the
absolute value or the negative
of this definite integral.
1
3
= −3 (x2 − 4x + 3)dx
1
3
1
A = − c a 13 × 33 − 2 × 32 + 3 × 3 b −
5 Evaluate the definite integral.
6 State the value of the
The area is
required area.
4
3
square units.
1
3
× 13 − 2 × 12 + 3 × 1 b d
PR
O
= 43
a
FS
square bracket notation.
− 2x + 3xd
2
O
A=
4 Perform the integration using
− c 13x3
b
a
c
A1
0
a
EC
b
c
a
b
x
A2
A2 = 3f(x)dx < 0, the required area is
b
c
b
TE
D
Since A1 = 3f(x)dx > 0 and
PA
G
E
Areas both above and below the x-axis
y
When dealing with areas that are both
above and below the x-axis, each area
must be evaluated separately.
y = f(x)
R
A = A1 + ∣ A2∣ = 3f(x)dx + 3 3f(x)dx 3
b
c
b
b
a
b
a
c
U
N
C
O
R
= A1 − A2 = 3f(x)dx − 3f(x)dx = 3f(x)dx + 3f(x)dx
WORKED
EXAMPLE
6
Find the area bounded by the curve y = x3 − 9x and the x-axis.
THINK
1 Factorise the cubic to find the x-intercepts.
WRITE/DRAW
y = x3 − 9x
y = x(x2 − 9)
y = x(x + 3)(x − 3)
The graph crosses the x-axis at x = 0
and x = ±3.
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 385
385
14/08/15 6:14 PM
y
12
10
8
6
4
2
y=
x3 – 9x
–5 –4 –3 –2 –1 0
–2
–4
–6
–8
–10
–12
1 2 3 4 5 x
FS
2 Sketch the graph, shading the required area.
3
PR
O
definite integral.
O
If we work out A = 3 (x3 − 9x)dx it comes to
3 The required area is given by a
−3
zero, as the positive and negative area have
cancelled out.
0
3
0
PA
G
−3
E
Let A1 = 3 (x3 − 9x)dx and A2 = 3 (x3 − 9x)dx,
so that A1 > 0 and A2 < 0, but A1 = ∣ A2 ∣ by
symmetry.
0
A1 = 3 (x3 − 9x)dx
TE
D
4 Perform the integration, using square
EC
bracket notation.
U
N
C
O
R
R
5 Evaluate the definite integral.
−3
A1 =
1 4
c 4x
A1 =
c a0
A1 = 81
4
0
−
9 2
x d
2
−3
− 14 × (−3) 4 − 92 × (−3) 2 b d
= 2014
3
A2 = 3 (x3 − 9x)dx = −81
= −2014
4
0
A1 + ∣ A2 ∣ = 2 × 81
= 81
= 4012
4
2
6 State the value of the required area.
The area is 4012 square units.
Area between curves
y
y2 = g(x)
If y1 = f(x) and y2 = g(x) are two continuous
curves that do not intersect between x = a
and x = b, then the area between the curves is
obtained by simple subtraction.
y1 = f(x)
0
386 a
b
x
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 386
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The area A1 is the entire shaded area bounded by the curve y2 = g(x), the x-axis and
b
the lines x = a and x = b, so A1 = 3g(x)dx. The red area, A2, is the area bounded by
AOS 3
Topic 2
b
the curve y1 = f(x), the x-axis and the lines x = a and x = b, so A2 = 3f(x)dx.
a
Concept 9
Areas of
bounded regions
Concept summary
Practice questions
a
The required area is the blue area, which is the area between the curves.
b
b
a
b
a
FS
A = A1 − A2 = 3g(x)dx − 3f(x)dx, and by the properties of definite integrals
A = A1 − A2 = 3 (g(x) − f(x))dx = 3 (y2 − y1)dx.
a
a
PR
O
Note that when finding areas between curves, it
does not matter if some of the area is above or
below the x-axis.
We can translate both curves up k units parallel
to the y-axis so that the area between the curves
lies entirely above the x-axis as shown below right.
O
b
y
y2 = g(x)
a
PA
G
E
0
b
b
y1 = f(x)
A = 3 (g(x) + k)dx − 3 ( f(x) + k)dx
a
b
a
TE
D
A = 3 (g(x) − f(x))dx
a
b
x
b
y
y2 = g(x) + k
EC
A = 3 ( y2 − y1)dx
a
U
N
C
O
R
R
Provided that y2 ≥ y1 for a < x < b, it does not matter
if some or all of the area is above or below the x-axis,
as the required area between the curves will be a
positive number. Note that only one definite integral is
required, that is y2 − y1 = g(x) − f(x). Evaluate this as
a one definite integral.
WORKED
EXAMPLE
7
a
0
x
b
Find the area between the parabola y = x2 − 2x − 15 and the straight line
y = 2x − 3.
THINK
1 Factorise the quadratic to find the
x-intercepts.
y1 = f(x) + k
WRITE/DRAW
y = x2 − 2x − 15
y = (x − 5)(x + 3)
The parabola crosses the x-axis at x = 5 and x = −3
and crosses the y-axis at y = −15.
The straight line crosses the x-axis at x = 32 and
crosses the y-axis at y = −3.
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 387
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14/08/15 6:14 PM
x2 − 2x − 15
x2 − 4x − 12
(x − 6)(x + 2)
x
= 2x − 3
=0
=0
= 6, −2
1 2 3 4 5 6 7 x
y = x 2 – 2x – 15
PA
G
–5 –4 –3 –2 –1 0
–2
–4
–6
–8
–10
–12
–14
y = 2x – 3
–16
–18
O
the straight line on one set of axes,
shading the required area.
FS
y
10
8
6
4
2
3 Sketch the graph of the parabola and
PR
O
intersection between the parabola and
the straight line.
Let y1 = x2 − 2x − 15 and y2 = 2x − 3.
To find the points of intersection,
solve y1 = y2.
E
2 Find the x-values of the points of
b
A = 3 (y2 − y1)dx with a = −2, b = 6,
4 The required area is given by a
TE
D
definite integral.
a
y1 = x2 − 2x − 15 and y2 = 2x − 3.
y2 − y1 = −x2 + 4x + 12
R
EC
6
O
bracket notation.
R
5 Perform the integration using square
U
N
C
6 Evaluate the definite integral.
7 State the value of the required
area between the parabola and
the straight line.
388 A = 3 (−x2 + 4x + 12)dx
−2
A=
1
c − x3
3
A=
1
×
3
1 1
c a−
−
a a3−×
3
A=
c a−
A = 8513
+
2x2
+ 12x d
6
−2
63 + 2 × 62 + 12 × 6 b
1
3 + 2 × (−2) 2 + 12 × (−
63 (−2)
+ 2 3×+622 +× 12
× 26+
× (−2)
b −
×
(−2)
12a a×−(−2)
bb d
3
The area between the straight line and the parabola
is 8513 square units.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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Exercise 8.2 Areas under and between curves
PRactise
Find the area bounded by the line y = 4x + 5, the x-axis, and the lines
x = 1 and x = 3. Check your answer algebraically.
3x
2 Find the area bounded by the line y = 4 −
and the coordinate axes. Check your
2
answer algebraically.
1
WE1
3
Find the area bounded by the curve y = 9 − x2, the x-axis and
the lines x = ±2.
WE2
4 The area bounded by the curve y = b − 3x2, the x-axis and the lines x = ±1 is
5
WE3
FS
equal to 16. Given that b > 3, find the value of b.
Find the area under one arch of the sine curve y = 4 sin(2x).
x
2
PR
O
7
O
6 Find the area under one arch of the curve y = 3 cosa b.
Find the area bounded by the coordinate axes, the graph of y = 6e3x and
the line x = 2.
WE4
8 Find the area bounded by the graph of y = 6(e−2x + e−2x), the x-axis and x = ±1.
WE5
Find the area bounded by the curve y = x2 − 5x + 6 and the x-axis.
E
9
11
WE6
PA
G
10 Find the area bounded by the curve y = 3x2 − 10x − 8 and the x-axis.
Find the area bounded by the curve y = x3 − 4x and the x-axis.
12 Find the area bounded by the curve y = 16x − x3, the x-axis, x = −2 and x = 4.
Find the area between the parabola y = x2 − 3x − 18 and the
straight line y = 4x − 10.
WE7
TE
D
13
14 Find the area corresponding to the region { y ≥ x2 − 2x − 8 } ∩ { y ≤ 1 − 2x } .
EC
15 a Find the area between the line y = 6 − 2x and the coordinate axis.
Check your answer algebraically.
b Find the area between the line y = 3x + 5, the x-axis, x = 1 and x = 4.
Check your answer algebraically.
R
Consolidate
U
N
C
O
R
16a Calculate the area bounded by:
ithe curve y = 12 − 3x2 and the x-axis
iithe curve y = 12 − 3x2, the x-axis and the lines x = ±1.
b Determine the area bounded by:
ithe graph of y = x2 − 25 and the x-axis
iithe graph of y = x2 − 25, the x-axis and the lines x = ±3.
17a Find the area under one arch of the sine curve y = 6 sina
πx
b.
3
πx
b.
2
c Find the area under one arch of the sine curve y = a sin(nx).
1
18a Find the area under the graph of y = between the x-axis and:
x
ix = 1 and x = 4
iix = 1 and x = e
iiix = 1 and x = a, where a > e.
b Find the area under one arch of the curve y = 4 cosa
Topic 8 Integral calculus Integral calculus 389
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b Find the area under the graph of y =
ix = 1 and x = 3
iix = 1 and x = a, where a > 1.
1
between the x-axis and:
x2
c Find the area bounded by the curve y = x2 − 2x − 15 and the x-axis.
19a Find the area bounded by the curve y = x2 + 3x − 18, the x-axis and the
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
lines x = −3 and x = 6.
b Find the area bounded by the curve y = x2 − 2x − 24, the x-axis and the
lines x = 2 and x = 8.
c Find the area bounded by:
ithe curve y = x3 − 36x and the x-axis
iithe curve y = x3 − 36x, the x-axis and the lines x = −3 and x = 6.
20a If a is a constant, find the area bounded by the curve y = x2 − a2 and
the x-axis.
b If a is a constant, find the area bounded by the curve y = x3 − a2x and
the x-axis.
21 Find the area between the curves:
a y = x2 and y = x
b y = x3 and y = x
c y = x4 and y = x
d y = x5 and y = x.
22 a i Find the area between the parabola y = x2 − 2x − 35 and the x-axis.
iiFind the area between the parabola y = x2 − 2x − 35 and the straight line
y = 4x − 8.
b i Find the area between the parabola y = x2 + 5x − 14 and the x-axis.
iiFind the area corresponding to the region
{ y ≥ x2 + 5x − 14 } ∩ { y ≤ 2x + 4 } .
2
23a Find the area between the line 2y + x − 5 = 0 and the hyperbola y = .
x
1
b Find the area between the line 9y + 3x − 10 = 0 and the hyperbola y = .
3x
24a Prove using calculus methods that the area of a right-angled triangle of base
length a and height b is given by 12ab.
O
b Prove using calculus that the area of a trapezium of side lengths a and b, and
U
N
C
h
width h is equal to (a + b).
2
x2
x
25 a Consider the graphs of y =
and y = 4 sina b.
Master
3
2
iFind the coordinates of the point of intersection between the graphs.
iiDetermine the area between the graphs, the origin and this point of
intersection, giving your answer correct to 4 decimal places.
−
x
4
x
and y = .
2
iFind the coordinates of the point of intersection between the graphs.
iiDetermine the area between the curves, the y-axis and this point of
intersection, giving your answer correct to 4 decimal places.
b Consider the graphs of y = 5e
390 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 390
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x
2
2x
b + 42 for x ≥ 0.
3
i Find the coordinates of the point of intersection between the graphs.
ii Determine the area between the curves, the y-axis and this point of
intersection, giving your answer correct to 4 decimal places.
190
x
b Consider the graphs of y =
− 5 and y = −32 cosa b for x ≥ 0.
2
5
x
i Find the coordinates of the first two points of intersection.
ii Determine the area between the curves and these first two points of
intersection, giving your answer correct to 4 decimal places.
O
Linear substitutions
A linear substitution is of the form u = ax + b.
PR
O
8.3
FS
26 a Consider the graphs of y = 23e and y = 45 sina
Finding integrals of the form 3 (ax + b) ndx where n ∈ Z
Integrals of the form 3 (ax + b) ndx, where a and b are non-zero real numbers and n is
AOS 3
E
Topic 2
Concept 5
8
Find 3 (2x − 5) 4dx.
EC
WORKED
EXAMPLE
TE
D
PA
G
a positive integer, can be performed using a linear substitution with u = ax + b.
du
The derivative
= a is a constant, and this constant factor can be taken outside the
dx
integral sign by the properties of indefinite and definite integrals. The integration
process can then be completed in terms of u. Note that since u has been introduced
in this solution process, the final answer must be given back in terms of the original
variable, x.
Antiderivatives
using linear
substitution
Concept summary
Practice questions
R
THINK
R
1 Although we could expand and integrate term by term,
C
O
it is preferable and easier to use a linear substitution.
U
N
2 Differentiate u with respect to x.
3 Express dx in terms of du by inverting both sides.
4 Substitute for dx.
5 Use the properties of indefinite integrals to transfer the
constant factor outside the front of the integral sign.
WRITE
Let u = 2x − 5.
4
4
3 (2x − 5) dx = 3u dx
u = 2x − 5
du
=2
dx
dx 1
=
du 2
dx = 12 du
4
41
3u dx = 3u 2du
= 123u4du
Topic 8 INTEGRAL CALCULUS
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391
14/08/15 6:15 PM
6 Perform the integration using 3undu =
un+1
n+1
1
1
2
× 15 u5 + c
1 5
= 10
u +c
with n = 4 so that n + 1 = 5, and add in the constant +c.
1
4
5
3 (2x − 5) dx = 10 (2x − 5) + c
7 Substitute back for u = 2x − 5 and express the final
answer in terms of x only and an arbitrary constant +c.
FS
Finding particular integrals of the form 3 (ax + b) ndx where n ∈ Q
O
Integrals of the form 3 (ax + b)ndx, where a and b are non-zero real numbers,
1
The gradient of a curve is given by
. Find the particular curve that
!4x + 9
passes through the origin.
E
9
PA
G
WORKED
EXAMPLE
PR
O
and n is a rational number, can also be performed using a linear substitution with
u = ax + b. First express the integrand (the function being integrated) as a power
using the index laws.
THINK
WRITE
dy
dx
.
TE
D
1 Recognise that the gradient of a curve is given by
2 Integrate both sides to give an expression for y.
3 Use index laws to express the integrand as a function
R
R
EC
to a power and use a linear substitution.
U
N
C
differentiate.
O
4 The integral cannot be done in this form, so
5 Express dx in terms of du by inverting both sides.
6 Substitute for dx.
dy
dx
=
1
!4x + 9
y=3
1
dx
!4x + 9
Let u = 4x + 9.
y = 3 (4x + 9) dx
−
1
2
y = 3u 2dx
−
u
du
dx
dx
du
dx
1
= 4x + 9
=4
1
4
1
= 4 du
=
y = 3u
− 12 1
du
4
y = 143u du
1
7 Use the properties of indefinite integrals to transfer
the constant factor outside the front of the integral sign.
392
−2
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 392
14/08/15 6:15 PM
y=
8 Perform the integration process using
1
n
n+1 with n = −1 so that n + 1 = 1 ,
3u du = n + 1 u
2
2
and add in the constant +c.
y=
y=
y=
9 Simplify and substitute for u = 4x + 9 to express the
answer in terms of x and an arbitrary constant +c.
1
4
1
1
2
× 1u + c
2
1
2 2
u +c
4
1
!u + c
2
1
!4x + 9
2
+c
be found using the given condition that the curve
passes through the origin.
c = −32
O
0 = 12 !0 + 9 + c
FS
Substitute y = 0 and x = 0 to find c:
10 The arbitrary constant +c in this particular case can
y = 12 !4x + 9 − 32
PR
O
11 Substitute back for c.
y = 12 (!4x + 9 − 3)
12 State the equation of the particular curve in a
factorised form.
E
Finding integrals of the form 3 (ax + b) ndx when n = −1
PA
G
Integrals of the form 3 (ax + b) ndx, when n = −1, a ≠ 0 and b ∈ R, involve the
1
logarithm function, since 3 dx = loge a ∣ x ∣ b + c.
x
10
1
.
5x + 4
TE
D
WORKED
EXAMPLE
Antidifferentiate
EC
THINK
R
1 Write the required integral.
O
R
2 Use a linear substitution.
U
N
C
3 The integral cannot be done in this form, so differentiate.
4 Express dx in terms of du by inverting both sides.
5 Substitute for dx.
6 Use the properties of indefinite integrals to transfer the
constant factor outside the front of the integral sign.
WRITE
1
35x + 4 dx
Let u = 5x + 4.
1
1
35x + 4 dx = 3u dx
u = 5x + 4
du
=5
dx
dx 1
=
du 5
1
dx = du
5
1
1 1
35x + 4 dx = 3u × 5 du
1 1
= 3u du
5
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 393
393
14/08/15 6:15 PM
1
1
35x + 4 dx = 5 loge a ∣ u ∣ b + c
7 Perform the integration process using 3 du = loge ∣u∣ + c.
u
1
1
1
35x + 4 dx = 5 loge a ∣5x + 4∣ b + c
8 Simplify and substitute for u = 5x + 4 to express the final
answer in terms of x only and an arbitrary constant +c, in
simplest form.
AOS 3
We can generalise the results from the last three examples to state:
n
3 (ax + b) dx =
Antiderivates
involving
logarithms
Concept summary
Practice questions
1
(ax + b) n+1 + c
a (n + 1)
•1
a
n ≠ −1
O
Concept 2
PR
O
Topic 2
FS
Finding integrals of the form 3 (ax + b) ndx where n ∈ Q
loge a ∣ ax + b ∣ b + c
n = −1
E
Evaluating definite integrals using a linear substitution
11
1
Evaluate 3
0
TE
D
WORKED
EXAMPLE
PA
G
When we evaluate a definite integral, the result is a number. This number is also
independent of the original variable used. When using a substitution, change the
terminals to the new variable and evaluate this definite integral in terms of the new
variable with new terminals. The following worked example clarifies this process.
4
dx.
(3x + 2) 2
EC
THINK
R
1 Write the integrand as a power using index laws and
C
O
R
transfer the constant factor outside the front of the
integral sign.
U
N
2 Use a linear substitution. Note that the terminals in the
definite integral refer to x-values.
3 The integral cannot be done in this form, so differentiate.
Express dx in terms of du by inverting both sides.
394
WRITE
1
3
0
4
dx
(3x + 2) 2
1
= 43 (3x + 2) −2dx
0
Let u = 3x + 2.
1
x=1
0
x=0
43 (3x + 2) −2dx = 4 3 u−2dx
u
du
dx
dx
du
dx
= 3x + 2
=3
1
3
1
= 3 du
=
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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When x = 0 ⇒ u = 2 and when
x = 1 ⇒ u = 5.
4 Change the terminals to the new variable.
u=5
4 3 u−2 13 du
5 Substitute for dx and the new terminals.
u=2
5
= 433u−2du
6 Transfer the constant factor outside the front of the
integral sign.
2
FS
5
2
PR
O
same value as the original definite integral. There
is no need to substitute back for x, and there is no
need for the arbitrary constant when evaluating a
definite integral.
4 1
c− d
3 u
O
=
7 The value of this definite integral has the
= 43 c −15 − a −12 b d
8 Evaluate this definite integral.
= 43 a 12 − 15 b
PA
G
E
− 2
= 43 a 5 10
b
1
4
2
3 (3x + 2) 2 dx = 5
0
TE
D
9 State the final result.
Finding integrals using a back substitution
U
N
C
O
R
R
EC
Integrals of the form 3x(ax + b) ndx can also be performed using a linear substitution
du
with u = ax + b. Since the derivative
= a is a constant, this constant can be taken
dx
outside the integral sign. However, we must express the integrand in terms of u only
1
before integrating. We can do this by expressing x in terms of u; that is, x = (u − b).
a
However, the final result for an indefinite integral must be given in terms of the
original variable, x.
WORKED
EXAMPLE
12
Find:
a 3x(2x − 5) 4dx
THINK
a 1 Use a linear substitution.
b 3
4x2
6x − 5
dx.
− 12x + 9
WRITE
a Let u = 2x − 5.
4
4
3x(2x − 5) dx = 3x u dx
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 395
395
14/08/15 6:15 PM
2 The integral cannot be done in this form,
so differentiate and express dx in terms
of du by inverting both sides.
u
du
dx
dx
du
dx
= 2x − 5
=2
1
2
1
= 2 du
=
2x = u + 5
3 Express x in terms of u.
1
4
41
3x(2x − 5) dx = 32 (u + 5)u 2 du
4 Substitute for x and dx.
6 Perform the integration, integrating term
final answer in terms of x only and an
arbitrary constant +c.
+ u5 b + c
1
1
4
6
5
3x(2x − 5) dx = 24 (2x − 5) + 4 (2x − 5) + c
u5
1 6 1 5
u + u + c = (u + 6) + c
24
4
24
TE
D
9 Alternatively, the result can be expressed
in a simplified form by taking out the
common factors.
EC
=
R
R
x only and an arbitrary constant +c.
O
b 1 Factorise the denominator as a
C
perfect square.
U
N
2 Use a linear substitution.
Differentiate and express dx in terms
of du by inverting both sides.
3 Express the numerator 6x − 5
in terms of u.
(2x − 5) 5
(2x − 5 + 6) + c
24
1
4
5
3x(2x − 5) dx = 24 (2x − 5) (2x + 1) + c
11 Express the final answer in terms of
396 1 6
u
6
PA
G
8 Substitute u = 2x − 5 and express the
and simplify.
a
1 6
= 24
u + 14u5 + c
7 Simplify the result by expanding.
10 Substitute back for u = 2x − 5
= 14 ×
E
by term and adding in the constant.
O
to transfer the constant factors outside the
front of the integral sign and expand the
integrand.
= 143 (u5 + 5u4)du
PR
O
5 Use the properties of indefinite integrals
FS
x = 12 (u + 5)
b 3
4x2
6x − 5
6x − 5
dx = 3
dx
− 12x + 9
(2x − 3) 2
Let u
du
dx
dx
du
dx
= 2x − 3.
=2
1
2
1
= 2 du
=
2x = u + 3
6x = 3(u + 3)
6x = 3u + 9
6x − 5 = 3u + 4
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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3u + 4
6x − 5
1
× du
3 (2x − 3) 2 dx = 3 u2
2
4 Substitute for 6x − 5, u and dx.
1 3u + 4
= 3a
bdu
2
u2
5 Use the properties of indefinite integrals
to transfer the constant factor outside the
front of the integral sign.
4
1 3
= 3 a + 2 bdu
2 u u
6 Simplify the integrand.
1 3
= 3 a u + 4u−2 bdu
2
1
= a 3 loge a ∣u∣ b − 4u−1 b + c
2
FS
7 Write in index form.
6x − 5
3
2
34x2 − 12x + 9 dx = 2 loge a ∣2x − 3∣ b − 2x − 3 + c
PA
G
answer in terms of x only and an arbitrary
constant +c, as before.
PR
O
9 Substitute u = 2x − 3 and express the final
1
4
= a3 loge a ∣u∣ b − u b + c
2
E
constant. The first term is a log, but in the
1
second term, we use 3undu =
un+1
n+1
with n = −2, so that n + 1 = −1.
O
8 Perform the integration, adding in the
Definite integrals using a back substitution
13
8
Evaluate 3
0
THINK
x
dx.
!2x + 9
TE
D
WORKED
EXAMPLE
WRITE
8
R
EC
1 Write the integrand as a power, using index laws. 3
C
O
R
2 Use a linear substitution,
U
N
3 The integral cannot be done in this form, so
differentiate. Express dx in terms of du by
inverting both sides.
0
8
x
dx = 3x(2x + 9) − 2dx
!2x + 9
0
Let u = 2x + 9.
8
3x(2x + 9)
1
−1
2
0
x=8
dx = 3 xu dx
−1
2
x=0
du
=2
dx
dx 1
=
du 2
dx = 12 du
4 Express x back in terms of u.
u = 2x + 9
2x = u − 9
x = 12 (u − 9)
5 Change the terminals to the new variable.
When x = 0, u = 9, and when x = 8, u = 25.
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 397
397
14/08/15 6:15 PM
x=8
3 xu dx
6 Substitute for dx, x and the new terminals.
x=0
−1
2
u=25
= 3 12 (u − 9)u
u=9
25
−1 1
2
du
2
= 14 3 (u − 9)u 2du
= 14 3 au2 − 9u
8 Expand the integrand.
1
9
3
−1
2
bdu
1 25
= 14 c 23u2 − 18u 2 d
9 Perform the integration.
3
FS
9
25
O
of the integral sign.
−1
PR
O
7 Transfer the constant factors outside the front
9
1
3
1
= 14 c a 23 (25) 2 − 18(25) 2 b − a 23 (9) 2 − 18(9) 2 b d
E
10 Evaluate the definite integral.
PA
G
2
2
= 14 c a 3 × 125 − 18 × 5 b − a 3 × 27 − 18 × 3 b d
8
x
22
3 !2x + 9 dx = 3
11 State the final result.
TE
D
0
Further examples involving back substitutions with logarithms
14
2x
Find 3
dx.
4x − 3
EC
WORKED
EXAMPLE
THINK
U
N
C
O
R
R
Method 1
1 Use a linear substitution. Differentiate
and express dx in terms of du by inverting
both sides.
2 Express the numerator 2x in terms of u.
WRITE
u
du
dx
dx
du
dx
= 4x − 3
=4
1
4
1
= 4 du
=
4x = u + 3
2x = 12 (u + 3)
3 Substitute for 2x, u and dx.
4 Use the properties of indefinite integrals to
transfer the constant factors outside the front
of the integral sign.
398
1
(u + 3) 1
2x
2
× du
dx
=
34x − 3
3 u
4
1 u+3
= 3 a u bdu
8
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 398
14/08/15 4:36 PM
3
1
= 3 a1 + u bdu
8
5 Simplify the integrand.
7 Substitute u = 4x − 3 and express the final
answer in terms of x only and an arbitrary
constant +c.
Method 2
1 Express the numerator as a multiple of the
denominator (in effect, use long division to
divide the denominator into the numerator).
1
a u + 3 loge a ∣u∣ b b + c
8
u 3
= + loge a∣u∣ b + c
8 8
4x − 3 3
2x
34x − 3 dx = 8 + 8 loge a ∣4x − 3∣ b + c
=
O
2x
1 4x
34x − 3 dx = 234x − 3 dx
FS
6 Perform the integration, adding in the constant.
PR
O
1 (4x − 3) + 3
= 3
dx
2
4x − 3
3
1
= 3 a1 +
bdx
2
4x − 3
3 Perform the integration, adding in the constant.
= 12 a x + 34 loge a ∣4x − 3∣ b b + c
E
2 Simplify the integrand.
PA
G
x 3
2x
34x − 3 dx = 2 + 8 loge a ∣4x − 3∣ b + c
4 State the final answer.
TE
D
5 Although the two answers do not appear
R
EC
to be the same, the log terms are identical.
4x − 3 x 3
However, since
= − , the two
8
2 8
answers are equivalent in x and differ in the
constant only, c1 = −38 + c.
4x − 3 3
+ loge a ∣4x − 3∣ b + c
8
8
x 3
= + loge a ∣4x − 3∣ b + c1
2 8
=
O
R
The situation above can often happen when evaluating indefinite integrals.
Answers may not appear to be identical, but after some algebraic or trigonometric
simplification, they are revealed to be equivalent and may differ by a constant only.
U
N
C
Exercise 8.3 Linear substitutions
PRactise
1
WE8
Find 3 (5x − 9) 6 dx.
2 Find 3 (3x + 4) 7dx.
3
1
A particular curve has a gradient equal to
. Find the particular
!16x
+
25
curve that passes through the origin.
WE9
4 Given that f ′(x) =
5
WE10
1
and f(2) = 3, find the value of f(1).
(3x − 7) 2
Antidifferentiate
1
.
3x − 5
Topic 8 Integral calculus Integral calculus 399
c08IntegralCalculus.indd 399
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Evaluate 3
WE11
−1
9
dx.
(2x + 3) 3
8 Find the area bounded by the graph of y =
axes and x = 4.
9
Find:
WE12
a 3x(5x − 9) 5dx
b 3
10 Find:
a 3
6
, the coordinate
!3x + 4
b 3
x
dx
(2x + 7) 3
5
0
1
−1
15x
dx.
(3x + 2) 2
PA
G
12 Evaluate 3
x
dx.
!3x + 1
13 WE14 Find 3
dx.
3x + 4
TE
D
6x
1
x
dx.
!6x + 5
E
11 WE13 Evaluate 3
2x − 1
dx.
9x − 24x + 16
2
FS
7
O
0
1
.
7 − 2x
PR
O
6 Find an antiderivative of
dx.
14 Evaluate 3
2x − 5
4x
Consolidate
EC
0
15 Integrate each of the following with respect to x.
R
R
a (3x + 5) 6
b
1
(3x + 5) 2
16 Find each of the following.
U
N
C
O
a 3 (6x + 7) 8dx
b 3
1
dx
!6x + 7
1
(3x + 5) 3
c 3
1
dx
6x + 7
d
a x(3x + 5) 6
b
x
(3x + 5) 2
c
x
(3x + 5) 3
a 3x(6x + 7) 8dx
b 3
c 3
d 3
x
dx
6x + 7
1
3
"
3x + 5
d 3
17 Integrate each of the following with respect to x.
18 Find each of the following.
400 c
d
1
dx
(6x + 7) 2
x
3
"
3x + 5
x
dx
!6x + 7
x
dx
(6x + 7) 2
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 400
14/08/15 4:36 PM
dx
1
=
and x(0) = 0, express x in terms of t.
dt (2 − 5t) 2
1
b A certain curve has its gradient given by
for x < 32. If the point
!3 − 2x
1
a − , −2 b lies on the curve, find the equation of the curve.
2
19a Given that
3
and that f(0) = 0, find f(1).
3 − 2x
x
d A certain curve has a gradient of
. Find the particular curve that passes
!2x
+
9
through the origin.
c Given that f ′(x) =
a 3 (3x − 4) 5dx
b 3x(3x − 4) 5dx
13
c 3
0
1
13
1
3
dt
"2t + 1
PR
O
1
2
O
2
FS
20 Evaluate each of the following.
d 3
0
t
3
"2t + 1
dt
21aSketch the graph of y = !2x + 1. Find the area between the curve, the
U
N
C
O
R
R
EC
TE
D
PA
G
E
coordinates axes and the line x = 4.
1
b Sketch the graph of y =
. Determine the area bounded by the curve, the
3x + 5
coordinate axes and x = 3.
1
c Sketch the curve y =
. Find the area bounded by this curve and the
(2x + 3) 2
x-axis between x = 1 and x = 2.
x
d Find the area bounded by the curve y =
, the coordinate
!16 − 3x
axes and x = 5.
22a Find the area of the region enclosed by the curves with the equations:
iy = 4 !x − 1 and y = 4 !3 − x
iiy2 = 16(x − 1) and y2 = 16(3 − x).
b Find the area between the curve y2 = 4 − x and the y-axis
c Determine the area of the loop with the equation y2 = x2 (4 − x).
di Find the area between the curve y2 = a − x where a > 0 and the y-axis
iiDetermine the area of the loop with the equation y2 = x2 (a − x), where a > 0.
23 Given that a and b are non-zero real constants, find each of the following.
x
1
dx
dx
a 3 !ax + b dx
b 3x!ax + b dx
c 3
d 3
ax + b
ax + b
Master
24 Given that a and b are non-zero real constants, find each of the following.
a 3
x
x
1
1
dx
dx
dx
dx
b 3
c 3
d 3
2
(ax + b)
(ax + b) 2
!ax + b
!ax + b
25 Given that a, b, c and d are non-zero real constants, find each of the following.
cx + d
cx2 + d
cx + d
cx2 + d
dx
dx
a 3
b 3
dx
c 3
dx
d 3
ax + b
ax + b
(ax + b) 2
(ax + b) 2
26 Given that a and b are non-zero real constants, find each of the following.
a 3
x2
dx
ax + b
b 3
x2
dx
(ax + b) 2
c 3
x2
dx
(ax + b) 3
d 3
x2
dx
!ax + b
Topic 8 Integral calculus Integral calculus 401
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8.4
Other substitutions
Non-linear substitutions
Topic 2
f(u)
3f(u)du
un n ≠ −1
un+1
n+1
1
u
eu
loge a ∣u∣ b
cos(u)
sin(u)
Concept 3
PR
O
eu
sin(u)
−cos(u)
sec2 (u)
tan(u)
15
Find 3
(x2
3x
dx.
+ 9) 2
THINK
WRITE
1 Write the integrand as a power using
index laws.
R
EC
2 Use a non-linear substitution.
R
3 The integral cannot be done in this form, so
U
N
C
O
differentiate. Express dx in terms of du by
inverting both sides.
4 Substitute for dx, noting that the terms
involving x will cancel.
5 Transfer the constant factors outside the front
of the integral sign.
6 The integral can now be done. Antidifferentiate
using 3undu =
un+1
with n = −2,
n+1
so that n + 1 = −1.
402
2
−2
33x(x + 9) dx
TE
D
WORKED
EXAMPLE
PA
G
E
Integration by
substitution
Concept summary
Practice questions
O
AOS 3
FS
The basic idea of a non-linear substitution is to reduce the integrand to one of the
standard u forms shown in the table below. Remember that after making a
substitution, x or the original variable should be eliminated. The integral must be
entirely in terms of the new variable u.
Let u = x2 + 9.
2
−2
−2
33x(x + 9) dx = 33xu dx
du
= 2x
dx
dx
1
=
du 2x
1
dx = du
2x
1
−2
−2
33xu dx = 33xu × 2xdu
= 323u−2du
= −32 u−1 + c
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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=−
7 Write the expression with positive indices.
8 Substitute back for x, and state the final result.
3
(x2
3
+c
2u
3
3x
dx = − 2
+c
2
+ 9)
2(x + 9)
Integrals involving the logarithm function
The result 3undu =
un+1
is true provided that n ≠ −1. When n = −1 we have the
n+1
Find 3
x2
O
16
x−3
dx.
− 6x + 13
THINK
WRITE/DRAW
E
Method 1
1 Use a non-linear substitution.
PR
O
WORKED
EXAMPLE
FS
1
special case 3 du = loge a ∣u∣b + c.
u
TE
D
PA
G
Let u = x2 − 6x + 13.
du
= 2x − 6
dx
= 2(x − 3)
dx
1
=
du 2(x − 3)
1
du
dx =
2(x − 3)
EC
2 Substitute for dx and u, noting that the
terms involving x cancel.
R
3 Transfer the constant outside the front
R
of the integral sign.
O
4 The integration can now be done.
U
N
C
1
Antidifferentiate using 3 du = loge a ∣u∣ b .
u
5 In this case, since
x2 − 6x + 13 = (x − 3) 2 + 4 > 0, for
all values of x, the modulus is not needed.
Substitute back for x, and state the
final result.
f ′(x)
d
Note that since [loge ( f(x)) ] =
,
dx
f(x)
f ′(x)
dx = loge a∣ f(x) ∣b + c.
it follows that 3
f(x)
x−3
x−3
1
3x2 − 6x + 13 dx = 3 u × 2(x − 3) du
1 1
= 3 u du
2
= 12 loge a ∣u∣ b + c
x−3
1
2
3x2 − 6x + 13 dx = 2 loge (x − 6x + 13) + c
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 403
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14/08/15 6:17 PM
2 Use the result 3
dx = loge a∣ f(x) ∣b + c,
f(x)
f ′ (x)
x−3
3x2 − 6x + 13 dx
=3
2(x − 3)
dx
2(x2 − 6x + 13)
2x − 6
1
= 3 2
dx
2 x − 6x + 13
x−3
1
2
3x2 − 6x + 13 dx = 2 loge (x − 6x + 13) + c
FS
Method 2
1 To make the numerator the derivative
of the denominator, multiply both the
numerator and the denominator by 2,
and take the constant factor outside the
front of the integral sign.
Examples involving trigonometric functions
O
with f(x) = x2 − 6x + 13.
PR
O
For trigonometric functions we use the results 3cos(u)du = sin(u) + c and
17
Find 3
THINK
1
1
cosa bdx.
2
x
x
PA
G
WORKED
EXAMPLE
E
3sin(u)du = −cos(u) + c.
WRITE
1
x
We choose this as the derivative of
1
1
is − 2,
x
x
O
R
R
EC
which is present in the integrand.
TE
D
1 Use a non-linear substitution. Let u = .
2 Substitute for u and dx, noting that the
U
N
C
x2 terms cancel.
3 Transfer the negative sign outside the integral sign.
4 Antidifferentiate, using 3cos(u)du = sin(u) + c.
5 Substitute back for x and state the answer.
404
1
u=x
= x −1
du
= −x −2
dx
1
=− 2
x
dx
= −x2
du
dx = −x2dx
3
1
1
cosa bdx
2
x
x
=3
1
cos(u) × −x2du
x2
= −3cos(u)du
= −sin(u) + c
1
1
1
3x2 cosa x bdx = −sina x b + c
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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examples involving exponential functions
For exponential functions we use the result 3eudu = eu + c.
18
Find:
a 3sin(2x) ecos(2x)dx
b
3
sin(!x)
tHINK
WrIte
a 1 Use a non-linear substitution. Let u = cos(2x).
a
O
u = cos(2x)
du
= −2 sin(2x)
dx
dx
−1
=
du 2 sin(2x)
−1
dx =
du
2 sin(2x)
PR
O
Choose this as the derivative of cos(2x) is −2 sin(2x),
which is present in the integrand. Note that the
substitution u = sin(2x) will not work.
dx.
!x
FS
WOrKeD
eXaMPle
3sin(2x)e
PA
G
E
2 Substitute for u and dx.
cos(2x)dx
= 3sin(2x)eu
= −123eudu
4 Antidifferentiate using 3eudu = eu + c.
= −12eu + c
TE
D
3 Transfer the constant factor outside the integral sign.
3sin(2x)e
EC
5 Substitute u = cos(2x) and state the answer.
b 1 Use a non-linear substitution. Let u = !x, but only
U
N
C
O
R
R
replace u in the numerator.
Express dx in terms of du by inverting both sides.
2 Substitute for u and dx, noting that the
!x terms cancel.
3 Transfer the constant factor outside the integral sign.
4 Antidifferentiate using 3sin(u)du = −cos(u) + c.
5 Substitute u = !x and state the answer.
cos(2x)
−1
du
2 sin(2x)
dx = −12ecos(2x) + c
1
b
u = !x = x 2
1
du 1 − 2
1
= x =
dx 2
2 !x
dx
= 2 !x
du
dx = 2 !x du
sin(!x)
1
3 (!x) dx = 3sin(!x) × !x dx
1
dx
= 3sin(u) ×
!x
= 23sin(u)du
= −2 cos(u) + c
3
sin(!x)
!x
dx = −2 cos(!x) + c
Topic 8 InTegral calculus
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14/08/15 4:36 PM
Definite integrals involving non-linear substitutions
When evaluating a definite integral, recall that the result is a number independent of
the original or dummy variable. In these cases, instead of substituting the original
variable back into the integral, change the terminals and work with the new definite
integral obtained.
Evaluate 3
0
THINK
t
"t2 + 4
dt.
FS
!5
WRITE
!5
3
1 Write the integrand as a power using
index laws.
0
!5
t
"t + 4
2
O
19
dt
PR
O
WORKED
EXAMPLE
2
2
3 t(t + 4) dt
−1
E
0
Let u = t2 + 4.
PA
G
2 Use a non-linear substitution.
!5
2
3 t(t + 4)
−1
2
t= !5
dt = 3 tu 2dt
0
−1
t=0
du
= 2t
dt
dt
1
=
du 2t
1
dt = du
2t
TE
D
3 The integral cannot be done in this form,
EC
so differentiate. Express dt in terms of du
by inverting both sides.
R
4 Change the terminals to the new variable.
R
t= !5
O
terminals, noting that the dummy
variable t cancels.
U
N
C
t=0
the front of the integral sign.
7 Perform the integration using 3undu =
n+1
un+1
406
−1
so that n + 1 =
1
.
2
−1
2
1
du
2t
u=4
9
6 Transfer the multiplying constant outside
with n =
u=9
2
3 tu dt = 3 tu
5 Substitute for dt and the new
−12,
When t = 0, u = 4, and when t = !5, u = 9.
= 123 u 2du
−1
4
1 9
= 12 c 2u 2 d
4
1 9
2
= cu d
4
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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= [ !9 − !4]
=3−2
=1
8 Evaluate the definite integral.
!5
3
9 State the answer.
0
t
"t + 4
2
dt = 1
Definite integrals involving inverse trigonometric functions
x
x
d
d
−1
1
, acos−1 a b b =
for a > 0 and
asin−1 a b b =
a
a
dx
"a2 − x2 dx
"a2 − x2
∣x∣ < a and d atan−1 a x b b = 2 a 2 for x ∈ R.
a
dx
a +x
20
1
2
cos−1 (2x)
dx.
Evaluate 3
!1 − 4x2
0
WRITE
PA
G
THINK
Let u = cos−1 (2x).
1 Use a non-linear substitution.
TE
D
1
2
2 The integral cannot be done in this form,
R
R
EC
so differentiate. Express dx in terms of du
by inverting both sides.
C
O
3 Change the terminals to the new variable.
U
N
E
WORKED
EXAMPLE
PR
O
O
FS
Recall that
4 Substitute for dx and the new terminals,
noting that the x terms cancel. Transfer the
constant multiple outside the front of the
integral sign.
3
0
x=
1
2
u
dx = 3
dx
"1 − 4x2
"1 − 4x2
cos − 1 (2x)
x=0
du
−2
=
dx "1 − 4x2
dx
= −12"1 − 4x2
du
dx = −12"1 − 4x2 du
When x = 12 , u = cos−1 (1) = 0, and
π
when x = 0, u = cos−1 (0) = .
2
x=
3
1
2
x=0
u
"1 − 4x2
=−
dx
u=0
1
1
u"1 − 4x2 du
3
2
2
"1 − 4x
u=π
2
0
= −12 3 u du
π
2
Topic 8 INTEGRAL CALCULUS
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14/08/15 6:18 PM
π
2
5 Using the properties of the definite integral,
swap the terminals to change the sign.
= 12 c 12u2 d
6 Perform the integration.
= 14 c u
1
2
3
8 State the answer.
Exercise 8.4 Other substitutions
2 Find 3
3
WE16
8x
dx.
+ 16) 3
"2x2 + 3
cos−1 (2x)
"1 − 4x2
dx =
π2
16
dx.
x2
x+2
dx.
+ 4x + 29
x2
dx.
x3 + 9
WE17
O
5
R
R
4 Find 3
(x2
5x
Find 3
2
TE
D
Find 3
0
EC
WE15
d
PA
G
0
1
0
π
2 2
π
= 14 c a b − 02 d
2
7 Evaluate the definite integral.
PRactise
π
2
FS
b
3 f 1 t 2 dt = − 3 f 1 t 2 dt
O
a
0
PR
O
a
E
b
= 123u du
Find 3
6 Find 3x cos(x2)dx.
C
U
N
1
1
sina bdx.
2
x
x
7
WE18
Find:
b 3
a 3cos(3x)esin(3x)dx
8a Find 3sec2 (2x)etan(2x)dx.
cos(!x)
!x
b Find 3
dx.
loge (3x)
4x
dx.
2!2
9
WE19
Evaluate 3
0
408 s
"2s2
+9
ds.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 408
14/08/15 6:18 PM
1
10 Evaluate 3
0
p
(3p2 + 5) 2
1
3
11 WE20 Evaluate 3
0
dp.
−1
sin (3x)
"1 − 9x2
dx.
FS
x
tan−1 a b
4
12 Evaluate 3
dx.
16 + x2
4
0
(x2
14a 3
c 3
d 3
x
dx
+ 9) 2
x2
dx
(x3 + 27) 3
x2
x3
+8
dx
4−x
"x2 − 8x + 25
dx
e2x
dx
4e2x + 5
EC
16 a 3
e−2x
dx
(3e−2x + 4) 3
R
c 3
TE
D
15 a 3 (x − 2)(x2 − 4x + 13) 3dx
c 3
U
N
C
O
R
17 a 3 sin(loge (4x))dx
c 3
1
x
x
sin−1 a b
2
"4 − x2
dx
b 3
E
c 3
b 3
x
dx
x +4
PR
O
13a 3x(x2 + 4) 5dx
O
For questions 13–17, find each of the indefinite integrals shown.
PA
G
Consolidate
2
x
"x2 + 9
dx
x2
"x3
+ 27
dx
d 3x2 (x3 + 8) 3dx
b 3
d 3
b 3
d 3
(x2
x2
x−2
dx
− 4x + 13) 2
4−x
dx
− 8x + 25
e−3x
dx
(2e−3x − 5) 2
2e2x + 1
dx
(e2x + x) 2
b 3 cos(loge (3x))dx
1
x
d 3
tan−1 (2x)
1 + 4x2
dx
18aA certain curve has a gradient given by x sin(x2). Find the equation of the
particular curve that passes through the origin.
dy 1
3
1
4
= 2 sec2 a b and when x = , y = 0, find y when x = .
b If
π
π
x
dx x
5−x
c Given that f ′(x) =
and f(0) = 0, find f(1).
2
x − 10x + 29
dy
π
= sin(2x)ecos(2x) and when x = , y = 0, find y when x = 0.
d If
dx
4
Topic 8 Integral calculus Integral calculus 409
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For questions 19–21, evaluate each of the definite integrals shown.
c 3
"t2 − 9
0
1
d 3
x
dx
(2x + 1) 2
2
1
3
3−x
20 a 3
dx
x2 − 6x + 34
b 3
4
2
4
π
8
b 3
E
loge (2t)
dt
3t
PA
G
1
2
4
d 3
1
1
cos a b dx
2
x
x
sin !x
dx
!x
1
TE
D
6
π
2−x
dx
(x − 4x + 5) 2
e
2
0
c 3
ds
2
0
21a 3sin(2 θ)ecos(2θ)d θ
3
π
"2s2 + 3
d 3sec2 (2θ ) etan (2θ) dθ
e!p
dp
!p
1
π
4
s
−1
5
c 3
x
dx
4x2 + 9
O
3
2
b 3
dt
PR
O
19 a 3
2
3t
FS
5
x
has vertical asymptotes at x = ±2. Find the value
x2 − c
of c and determine the area bounded by the curve, the x-axis and the
lines x = 3 and x = 5.
b Find the area bounded by the curve y = x cos(x2), the x-axis and the
!π
.
lines x = 0 and x =
2
2
c Find the area bounded by the curve y = xe−x , the x-axis and the
lines x = 0 and x = 2.
x
d Find the area bounded by the curve y =
, the x-axis and the
2 + 4
"x
lines x = 0 and x = 2 !3.
U
N
C
O
R
R
EC
22 a
The graph of y =
Master
23 If a, b ∈ R\ { 0 } , then find each of the following.
a 3
x
"ax2
+b
b 3
dx
c 3x(ax2 + b) ndx
d 3
n ≠ −1
(ax2
ax2
x
dx
+ b) 2
x
dx
+b
24 Deduce the following indefinite integrals, where f(x) is any function of x.
a 3
410 f ′(x)
f(x)
dx
b 3
f ′(x)
(f(x))
dx
2
c 3
f ′(x)
!f(x)
dx
d 3f ′(x)e f(x)dx
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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8.5
Integrals of powers of trigonometric
functions
Introduction
In this section we examine indefinite and definite integrals involving powers of
Concept 6
Antiderivatives
with
trigonometric
identities
Concept summary
Practice questions
PR
O
Integrals involving sin2(kx) and cos2(kx)
The trigonometric double-angle formulas
(1) sin(2A) = 2 sin(A)cos(A)
and
(2) cos(2A) = cos2 (A) − sin2 (A)
O
Topic 2
FS
trigonometric functions of the form 3sinm (kx)cosn (kx)dx where m, n ∈ J.
AOS 3
= 2 cos2 (A) − 1
= 1 − 2 sin2 (A)
(4)
PA
G
E
are useful in integrating certain powers of trigonometric functions.
Rearranging (2) gives
(3) sin2 (A) = 12 (1 − cos(2A)) and
cos2 (A) = 12 (1 + cos(2A))
Find:
21
a 3cos2 (3x)dx
EC
WORKED
EXAMPLE
TE
D
To integrate sin2 (kx) use (3); to integrate cos2 (kx) use (4).
THINK
1
(1
2
R
a 1 Use the double-angle formula
C
O
R
cos2 (A) =
+ cos(2A)) with A = 3x.
Transfer the constant factor outside the front
of the integral sign.
2 Integrate term by term,
U
N
using 3cos(kx) =
1
sin(kx) + c with k = 6.
k
3 Expand and state the final result.
b 1 Use the double-angle formula
2 sin(A)cos(A) = sin(2A) with A = 3x,
so that sin2 (A)cos2 (A) = 14 sin2 (2A).
b 3sin2 (3x)cos2 (3x)dx.
WRITE
a 3cos2 (3x)dx
= 123 (1 + cos(6x))dx
= 12 c x + 16 sin(6x) d + c
x
1
2
3cos (3x)dx = 2 + 12 sin(6x) + c
b 3sin2 (3x)cos2 (3x)dx
= 143 (2 sin(3x)cos(3x)) 2dx
= 143sin2 (6x)dx
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 411
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14/08/15 4:38 PM
= 14312 (1 − cos(12x))dx
2 Use the double-angle formula
sin2 (A) = 12 (1 − cos(2A)) with A = 6x.
= 183 (1 − cos(12x))dx
1
= 18 c x − 12
sin(12x) d + c
3 Integrate term by term, using
1
3cos(kx) = k sin(kx) + c with k = 12.
x
1
2
2
3sin (3x)cos (3x)dx = 8 − 96 sin(12x) + c
FS
4 Expand and state the final result.
PR
O
O
Note that as well as the double-angle formulas, there are many other relationships
between trigonometric functions, for example sin2 (A) + cos2 (A) = 1.
Often answers to integrals involving trigonometric functions can be expressed in
several different ways, for example, as powers or multiple angles. Answers derived
from CAS calculators may appear different, but often they are actually identical and
differ in the constant term only.
22
Find 3sin(3x)cos4 (3x)dx.
THINK
TE
D
WORKED
EXAMPLE
PA
G
E
Integrals involving sin(kx)cosm (kx) and cos(kx)sinm (kx) where m > 1
Integrals of the forms sin(kx)cosm (kx) and cos(kx)sinm (kx) where m > 1 can be
performed using non-linear substitution, as described in the previous section.
WRITE
1 Use a non-linear substitution. Let
O
R
R
EC
u = cos(3x). We choose this as the derivative
of cos(3x) is −3 sin(3x), which is present in
the integrand.
C
2 Substitute for u and dx, noting that the
U
N
x terms cancel.
3 Transfer the constant factor outside the
integral sign.
4 Antidifferentiate.
u = cos(3x)
du
= −3 sin(3x)
dx
dx
−1
=
du 3 sin(3x)
−1
dx =
du
3 sin(3x)
4
3sin(3x)cos (3x)dx
= 3sin(3x)u4 ×
= −133u4du
−1
du
3 sin(3x)
1 u5
=− ×
+c
3 5
1 5
= −15
u +c
5 Substitute back for x and state the final result.
412
1
4
5
3sin(3x)cos (3x)dx = −15 cos (3x) + c
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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Integrals involving odd powers
Antidifferentiation of sinm (kx)cosn (kx) when at least one of m or n is an odd
power can be performed using a non-linear substitution and the formula
sin2 (A) + cos2 (A) = 1.
WORKED
EXAMPLE
23
Find an antiderivative of sin3 (2x)cos4 (2x).
THINK
WRITE
FS
3
4
3sin (2x)cos (2x)dx
1 Write the required antiderivative.
2
4
3sin(2x)sin (2x)cos (2x)dx
2 Break the odd power.
sin3 (2x) = sin(2x)sin2 (2x), and
sin2 (A) = 1 − cos2 (A).
PR
O
O
= 3sin(2x)(1 − cos2 (2x))cos4 (2x)dx
u = cos(2x)
du
= −2 sin(2x)
dx
dx
−1
=
du 2 sin(2x)
−1
dx =
du
2 sin(2x)
3 Use a non-linear substitution. Let u = cos(2x).
PA
G
E
We choose this as the derivative of cos(2x) is
−2 sin(2x), which is present in the integrand and
will cancel.
= 3sin(2x)(1 − u2)u4 ×
4 Substitute for u and dx, noting that the sin(2x)
TE
D
terms cancel.
= −123 (1 − u2)u4du
5 Transfer the constant factor outside the integral
R
= −12 ×
O
C
U
N
24
1 5
u
5
− 17u7 b + c
3
4
3sin (2x)cos (2x)dx
7 Substitute back for x and state the final result.
WORKED
EXAMPLE
a
1 7
1 5
u − 10
u +c
= 14
R
6 Antidifferentiate.
= −123 (u4 − u6)du
EC
sign and expand.
−1
du
2 sin(2x)
1
1
= 14
cos7 (2x) − 10
cos5 (2x) + c
Integrals involving even powers
Antidifferentiation of sinm (kx)cosn (kx) when both m and n are even powers must be
performed using the double-angle formulas.
Find an antiderivative of cos4 (2x).
THINK
1 Write the required antiderivative.
WRITE
4
3cos (2x)dx
Topic 8 INTEGRAL CALCULUS
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2 Since there is no odd power, we
must use the double-angle formula
cos2 (A) = 12 (1 + cos(2A)) with A = 2x.
2
2
3 (cos (2x)) dx
= 3 a 12 (1 + cos(4x)) b dx
2
= 143 (1 + 2 cos(4x) + cos2 (4x))dx
3 Expand the integrand
4 Replace cos2 (A) = 2 (1 + cos(2A)) with
= 143 a 1 + 2 cos(4x) + 12 (1 + cos(8x)) b dx
5 Antidifferentiate term by term.
=
1
FS
= 143 a 32 + 2 cos(4x) + 12 cos(8x) b dx
1 3x 2
1
sin(8x) d + c
c + sin(4x) +
4 2
4
16
O
A = 4x and expand. The integrand is now
in a form that we can integrate term by term.
PR
O
3x 1
1
4
3cos (2x)dx = 8 + 8 sin(4x) + 64 sin(8x) + c
6 State the final result.
E
Integrals involving powers of tan (kx)
PA
G
In this section we examine indefinite and definite integrals involving powers of the
25
Find 3tan(2x)dx.
EC
WORKED
EXAMPLE
THINK
R
sin(A)
with A = 2x.
cos(A)
R
1 Use tan(A) =
TE
D
tangent function, that is, integrals of the form 3tann (kx)dx where n ∈ J.
sin(A)
The result tan(A) =
is used to integrate tan(A).
cos(A)
O
2 Use a non-linear substitution. Let u = cos(2x).
U
N
C
We choose this as the derivative of cos(2x) is
−2 sin(2x), which is present in the numerator
integrand and will cancel.
3 Substitute for u and dx, noting that the x
terms cancel.
4 Transfer the constant factor outside the
integral sign.
414
WRITE
sin(2x)
3tan(2x)dx = 3cos(2x) dx
u = cos(2x)
du
= −2 sin(2x)
dx
dx
−1
=
du 2 sin(2x)
−1
dx =
du
2 sin(2x)
3tan(2x)dx = 3
sin(2x)
−1
×
du
u
2 sin(2x)
1 1
= − 3u du
2
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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= −12 loge a ∣u∣b + c
5 Antidifferentiate.
6 Substitute back for x and state the final result.
Again note that there are different answers
using log laws and trigonometric identities.
1
3tan(2x)dx = −2 loge a ∣cos(2x) ∣b + c
= 12 loge a ∣cos(2x) ∣b
−1
+c
= 12 loge a ∣sec(2x) ∣b + c
Integrals involving tan2 (kx)
26
Find 3tan2 (2x)dx.
WRITE
1 Use 1 + tan2 (A) = sec2 (A) with A = 2x.
2 Use 3sec2 (kx) = tan(kx) + c and integrate
k
TE
D
1
O
2
2
3tan (2x)dx = 3 (sec (2x) − 1)dx
PA
G
THINK
E
WORKED
EXAMPLE
PR
O
d
1
(tan(kx)) = k sec2 (kx), so that 3sec2 (kx) = tan(kx) + c.
dx
k
FS
To find 3tan2 (kx)dx, we use the results 1 + tan2 (A) = sec2 (A) and
1
2
3tan (2x)dx = 2 tan(2x) − x + c
EC
term by term. State the final result.
EXERCISE 8.5 Integrals of powers of trigonometric functions
WE21
Find:
a 3sin2 a bdx
4
b 3sina bcosa bdx.
4
4
a 34 sin2 (2x)dx
b 3sin2 (2x)cos2 (2x)dx.
R
1
x
R
PRACTISE
U
N
C
O
2 Evaluate:
π
6
0
3
WE22
Find 3cos(4x)sin5 (4x)dx.
x
x
π
3
0
π
4
4 Evaluate 3sin(2x)cos3 (2x)dx.
0
5
WE23
Find an antiderivative of cos5 (4x)sin2 (4x).
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 415
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14/08/15 4:38 PM
π
12
6 Evaluate 3 sin3 (3x)dx.
0
7
WE24
Find an antiderivative of sin4 (2x).
π
12
8 Evaluate 3 cos4 (3x)dx.
0
WE25
x
Find 3 tana bdx.
2
FS
9
π
12
PR
O
O
10 Evaluate 3 tan(4x)dx.
0
11 WE26 Find 3tan2 a bdx.
3
x
0
Consolidate
13 Find each of the following.
a 3cos(2x)sin(2x)dx
PA
G
12 Evaluate 3 tan2 (4x)dx.
E
π
16
TE
D
c 3cos3 (2x)sin(2x)dx
b 3 (cos2 (2x) + sin2 (2x))dx
d 3cos(2x)sin3 (2x)dx
14 Evaluate each of the following.
EC
π
4
R
0
π
4
R
a 3cos(2x)sin4 (2x)dx
U
N
C
O
c 3cos2 (2x)sin2 (2x)dx
0
15 Find each of the following.
π
4
b 3cos2 (2x)sin3 (2x)dx
0
π
4
d 3cos3 (2x)sin2 (2x)dx
0
a 3cos2 (4x)sin2 (4x)dx
b 3cos2 (4x)sin3 (4x)dx
c 3cos3 (4x)sin2 (4x)dx
d 3cos3 (4x)sin4 (4x)dx
16 Find an antiderivative of each of the following.
a cosec2 (2x)cos(2x)
c
416 sin(2x)
cos3 (2x)
b sec2 (2x)sin(2x)
d
cos(2x)
sin3 (2x)
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 416
14/08/15 4:38 PM
17 Evaluate each of the following.
π
6
π
6
a 3sin2 (3x)dx
b 3cos3 (3x)dx
c 3sin4 (3x)dx
d 3cos5 (3x)dx
0
π
6
0
π
6
0
0
FS
18 a Find 3 (cos(2x) + sin(2x)) 2dx.
O
b Find 3cos3 (2x) + sin3 (2x)dx.
19 Find each of the following.
PA
G
a 3tan(3x)dx
E
ia double-angle formula
iithe substitution u = cos(2x)
iiithe substitution u = sin(2x).
PR
O
c Consider 3sin3 (2x)cos3 (2x)dx. Show that this integration can be done using:
TE
D
c 3tan(3x)sec2 (3x)dx
b 3cot(3x)dx
d 3tan2 (3x)sec2 (3x)dx
20 Evaluate each of the following.
π
20
0
C
O
R
c 3 tan3 (5x)sec2 (5x)dx
U
N
b 3 tan2 (5x)dx
0
R
π
20
EC
a 3 tan(5x)dx
π
20
0
π
20
d 3 tan2 (5x)sec4 (5x)dx
0
21 Given that n ≠ −1 and a ∈ R, find each of the following.
a 3sin(ax)cosn (ax)dx
b 3cos(ax)sinn (ax)dx
c 3tann (ax)sec2 (ax)dx
22 Integrate each of the following where a ∈ R\{0}.
a tan(ax)
Master
b tan2 (ax)
c tan3 (ax)
d tan4 (ax)
23 Integrate each of the following where a ∈ R\ { 0 } .
a sin2 (ax)
b sin3 (ax)
c sin4 (ax)
d sin5 (ax)
24 Integrate each of the following where a ∈ R\ { 0 } .
a cos2 (ax)
b cos3 (ax)
c cos4 (ax)
d cos5 (ax)
Topic 8 Integral calculus
c08IntegralCalculus.indd 417
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involving inverse trigonometric
8.6 Integrals
functions
Integrals involving the inverse sine function
x
x
d
1
1
, it follows that 3
dx = sin−1 a b + c for
asin−1 a b b =
a
a
dx
"a2 − x2
"a2 − x2
a > 0 and ∣x∣ < a.
Since
tHinK
"36 −
x2
b 3
dx
4
"49 −
a
1
x
dx = sin−1 a b + c with a = 6.
a
"a2 − x2
3
12
"36 − x2
= 123
3
O
C
U
N
1
−
u2
u
du = sin−1 a b + c with a = 7.
a
5 Substitute back for u and express the final answer in
terms of x only and an arbitrary constant +c.
dx
4
"49 − 36x2
dx
1
× du
"49 − u2 6
2
1
= 3
du
3 "49 − u2
constant factor outside the front of the integral sign.
"a2
"36 − x2
=3
3 Use the properties of indefinite integrals to transfer the
4 Use 3
1
dx = 16 du
R
R
EC
TE
D
express dx in terms of du by inverting both
sides.
dx
x
= 12 sin−1 a b + c
6
b u = 6x
u2 = 36x2
du
=6
dx
dx 1
=
du 6
PA
G
b 1 Use a linear substitution with u = 6x and
2 Substitute for dx and u.
dx.
36x2
WritE
E
a Use 3
12
O
a 3
FS
Find:
27
PR
O
WOrKeD
eXaMPLe
=
3
4
2 −1 u
sin a b + c
7
3
4
"49 − 36x2
dx =
2 −1 6x
sin a b + c
7
3
Integrals involving the inverse cosine function
x
x
d
−1
−1
, it follows that 3
dx = cos−1 a b + c
acos−1 a b b =
a
a
dx
"a2 − x2
"a2 − x2
where a > 0 and ∣x∣ < a.
Since
418
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c08IntegralCalculus.indd 418
14/08/15 4:38 PM
WORKED
EXAMPLE
−4
On a certain curve the gradient is given by
. Find the equation of
the curve that passes through the origin. "81 − 25x2
28
WRITE
dy
dx
.
2 Integrating both sides gives an expression for y.
3 Use a linear substitution with u = 5x and express dx
−4
"81 − 25x2
u = 5x
u2 = 25x2
du
=5
dx
dx 1
=
du 5
dx = 15 du
y=3
E
4 Substitute for dx and u.
dx
constant factor outside the front of the integral sign.
−1
TE
D
u
du = cos−1 a b + c with a = 9.
a
"a2 − u2
7 Substitute back for u and express the final answer
EC
in terms of x only and an arbitrary constant +c.
8 Since the curve passes through the origin, we can
O
R
R
let y = 0 when x = 0 to find c.
U
N
C
9 Substitute back for c.
10 State the equation of the particular curve in a
factored form.
11 Note that an alternative answer is
π
possible, since sin−1 (x) + cos−1 (x) = .
2
−4
"81 −
u2
1
× du
5
−1
4
y= 3
du
5 "81 − u2
PA
G
5 Use the properties of indefinite integrals to transfer the
6 Use 3
y=3
PR
O
in terms of du by inverting both sides.
dy
−4
=
dx "81 − 25x2
FS
1 Recognise that the gradient of a curve is given by
O
THINK
y=
y=
u
4
cos−1 a b + c
9
5
5x
4
cos−1 a b + c
9
5
4
cos−1 (0) + c
5
4 π
0= × +c
5 2
2π
c=−
5
5x
2π
4
y = cos−1 a b −
5
9
5
0=
5x
π
4
y = acos−1 a b − b
5
9
2
5x
4
y = − sin−1 a b
9
5
Integrals involving the inverse tangent function
Since
a
x
x
d
1
1
, it follows that 3 2
= tan−1 a b + c for x ∈ R.
atan−1 a b b = 2
2
2
a
a
a
dx
a +x
a +x
Topic 8 INTEGRAL CALCULUS
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419
14/08/15 6:19 PM
29
Antidifferentiate each of the following with respect to x.
a
12
36 + x2
b
4
49 + 36x2
THINK
WRITE
dx
a 3
36 + x2
12
123
1
dx
36 + x2
x
1
= 12 × tan−1 a b + c
6
6
x
= 2 tan−1 a b + c
6
x
1
1
= tan−1 a b + c with a = 6.
2
a
a
+x
dx
b 3
49 + 36x2
4
b 1 Write the required antiderivative.
TE
D
express dx in terms of du by inverting
both sides.
EC
3 Substitute for dx and u.
4 Use the properties of indefinite integrals to
R
x
1
1
= tan−1 a b + c with a = 7.
2
a
a
+x
O
a2
C
5 Use 3
= 6x
= 36x2
=6
1
6
1
= 6 du
=
4
4
1
349 + 36x2 dx = 349 + u2 × 6 du
2
1
= 3
du
3 49 + u2
R
transfer the constant factor outside the front
of the integral sign.
u
u2
du
dx
dx
du
dx
PA
G
2 Use a linear substitution with u = 6x and
FS
a2
O
2 Use 3
PR
O
a 1 Write the required antiderivative.
E
WORKED
EXAMPLE
U
N
6 Substitute back for u and express the final
answer in terms of x only and an arbitrary
constant +c.
=
u
2 1
× tan−1 a b + c
7
3 7
4
2
−1 6x
349 + 36x2 dx = 21 tan a 7 b + c
Definite integrals involving inverse trigonometric functions
WORKED
EXAMPLE
30
Evaluate:
2
9
a 3
−2
0 "16
420
− 81x2
4
9
dx
dx.
b 3
16 + 81x2
2
0
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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14/08/15 6:19 PM
THINK
WRITE
a 1 Use a linear substitution with u = 9x and
a
express dx in terms of du by inverting
both sides.
u
u2
du
dx
dx
du
dx
= 9x
= 81x2
=9
1
9
=
= 19 du
0
−2
"16 − 81x2
2
= 3
0
−2
1
× du
"16 − u2 9
2
front of the integral sign.
−1
E
0
5 Perform the integration using
u
du = cos−1 a b + c with a = 4.
a
"a2 − u2
TE
D
3
2
−1
= 3
du
9 "16 − u2
PA
G
4 Transfer the constant factor outside the
dx
PR
O
3
3 Substitute for dx, u and the new terminals.
O
2
9
FS
When x = 29 , u = 2, and when x = 0, u = 0.
2 Change the terminals to the new variable.
EC
6 Evaluate the definite integral.
R
R
O
3
C
7 State the final result. Note that since this is just
U
N
evaluating a definite integral and not finding an
area, the answer stays as a negative number.
b 1 Use a linear substitution with u = 9x and
express dx in terms of du by inverting
both sides.
2 Change the terminals to the new variable.
u
2
c cos−1 a b d
9
4 0
=
2
1
c cos−1 a b − cos−1 (0) d
9
2
2 π π
c − d
9 3 2
2 π(2 − 3)
= a
b
9
6
=
2
9
0
b
2
=
−2
"81 − 16x2
u
u2
du
dx
dx
du
dx
dx = −
π
27
= 9x
= 81x2
=9
1
9
= 19 du
=
When x = 49 , u = 4, and when x = 0, u = 0.
Topic 8 Integral calculus Integral calculus 421
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14/08/15 6:20 PM
4
9
2
3 16 + 81x2 dx
3 Substitute for dx, u and the new terminals.
0
4
1
2
=3
× du
2
9
16 + u
0
4
front of the integral sign.
FS
2
1
= 3
du
9 16 + u2
4 Transfer the constant factor outside the
0
3
x
1
1
= tan−1 a b + c with a = 4.
2
a
a
a +x
2
=
1 3 −1
tan (1) − tan−1 (0) 4
18
1 π
=
c − 0d
18 4
=
PA
G
E
6 Evaluate this definite integral.
4
u
1
c tan−1 a b d
18
4 0
PR
O
5 Perform the integration using
O
4
u
2 1
= c tan−1 a b d
9 4
4 0
4
9
π
2
3 16 + 81x2 dx = 72
7 State the final result.
TE
D
0
R
R
EC
Integrals involving completing the square
A quadratic expression in the form ax2 + bx + c can be expressed in the form
a(x + h) 2 + k (the completing the square form). This can be used to integrate
1
expressions of the form 2
with a > 0 and Δ = b2 − 4ac < 0, which
ax + bx + c
will then involve the inverse tangent function.
O
Integrating expressions of the form
1
U
N
C
+ bx + c
will involve the inverse sine or cosine functions.
WORKED
EXAMPLE
31
with a < 0 and Δ = b2 − 4ac > 0
Find each of the following with respect to x.
a 3
9x2
b 3
1
dx
+ 12x + 29
THINK
a 1 Express the quadratic factor in the
denominator in the completing the square
form by making it into a perfect square.
422
"ax2
WRITE
1
"21 − 12x − 9x2
dx
a 9x2 + 12x + 29
= (9x2 + 12x + 4) + 25
= (3x + 2) 2 + 25
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 422
14/08/15 6:20 PM
1
39x2 + 12x + 29 dx
2 Write the denominator as the sum of
two squares.
=3
and express dx in terms of du by inverting
both sides.
FS
3 Use a linear substitution with u = 3x + 2
1
dx
(3x + 2) 2 + 25
u = 3x + 2
du
=3
dx
dx 1
=
du 3
dx = 13 du
1
1
1
3 (3x + 2) 2 + 25 dx = 3u2 + 25 × 3 du
O
4 Substitute for dx and u.
PR
O
1
1
du
= 3 2
3 u + 25
5 Transfer the constant factor outside the
front of the integral sign.
u
1
1
= tan−1 a b + c with a = 5.
2
a
a
a +u
=
E
2
=
PA
G
6 Use 3
7 Substitute back for u and express the final
TE
D
answer in terms of x only and an arbitrary
constant +c.
b 1 Express the quadratic factor in the
EC
denominator in the completing the square
form by making it into the difference of
two squares.
R
2 Write the denominator as the difference of
O
C
3 Use a linear substitution with u = 3x + 2
U
N
and express dx in terms of du by inverting
both sides.
u
1
tan−1 a b + c
15
5
1
1
−1 3x + 2
39x2 + 12x + 29 dx = 15 tan a 5 b + c
b 21 − 12x − 9x2
3
= 21 − (9x2 + 12x)
= 21 − (9x2 + 12x + 4) + 4
= 25 − (3x + 2) 2
1
"21 − 12x − 9x2
=3
R
two squares under the square root.
u
1 1
× tan−1 a b + c
3 5
5
dx
1
"25 − (3x + 2) 2
dx
u = 3x + 2
du
=3
dx
dx 1
=
du 3
dx = 13 du
4 Substitute for dx and u.
3
1
"25 − (3x + 2) 2
=3
1
"25 −
u2
dx
1
× du
3
Topic 8 Integral calculus Integral calculus 423
c08IntegralCalculus.indd 423
14/08/15 5:19 PM
1
1
du
= 3
3 "25 − u2
5 Transfer the constant factor outside the front
of the integral sign.
6 Use 3
1
u
= sin−1 a b + c with a = 5.
a
"a2 − u2
=
3
7 Substitute back for u and express the final
1
"21 − 12x − 9x
2
1 −1 3x + 2
sin a
b+c
3
5
dx =
FS
answer in terms of x only and an arbitrary
constant +c.
1 −1 u
sin a b + c
3
5
O
Integrals involving substitutions and inverse trigonometric functions
We can break up complicated integrals into two manageable integrals using the
Find 3
THINK
4x + 5
"25 − 16x2
dx.
E
32
PA
G
WORKED
EXAMPLE
PR
O
following property of indefinite integrals: 3 ( f(x) ± g(x))dx = 3f(x)dx ± 3g(x)dx.
WRITE
1 If the x was not present in the numerator, the integral
EC
O
R
R
2 Write the first integral as a power using index laws.
C
3 Use a non-linear substitution with u = 25 − 16x2 and
U
N
express dx in terms of du by inverting both sides.
4 Substitute for dx and u, noting that the
x terms cancel.
424
4x + 5
"25 − 16x2
=3
TE
D
would involve an inverse sine function. If the 5 was
not present in the numerator, the integral would
involve a non-linear substitution. Break the integral
into two distinct problems: one involving a nonlinear substitution and one involving the inverse
trigonometric function.
3
3
dx
4x
"25 − 16x2
4x
"25 − 16x2
dx + 3
5
"25 − 16x2
dx
dx
= 34x(25 − 16x2)
−1
2
dx
u = 25 − 16x2
du
= −32x
dx
dx
1
=−
du
32x
1
dx = −
du
32x
2
34x(25 − 4x )
= 34xu
−1
2
−1
2
dx
×−
1
du
32x
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 424
14/08/15 6:20 PM
−1
1
−83u 2du
5 Transfer the constant factor outside the integral sign.
=
6 Perform the integration using 3undu =
un+1 with
n+1
1
= −18 × 1 u2
1
3
7 Substitute back for u.
8 Consider the second integral. Use a linear substitution
1
dx = − "25 − 16x2
4
"25 − 4x
x
2
5
"25 − 16x2
v = 4x
dv
=4
dx
dx 1
=
dv 4
1
dx = dv
4
dx
E
PR
O
with v = 4x and express dx in terms of dv by inverting
both sides.
3
=
2
1
−4 !u
FS
n = −12, so that n + 1 = 12.
O
1
PA
G
3
10 Use the properties of indefinite integrals to transfer the
constant factor outside the front of the integral sign.
v
dv = sin−1 a b + c with a = 5.
a
"a2 − v2
1
R
R
12 Substitute back for v.
EC
11 Use 3
O
13 Express the original integral as the sum as
U
N
C
the two integrals, adding in only one arbitrary
constant + c.
5
"25 − 15x2
=3
dx
5
"25 −
v2
1
× dv
4
5
1
= 3
dv
4 "25 − v2
TE
D
9 Substitute for dx and v.
=
3
3
5 −1 v
sin a b
4
5
5
"25 −
4x2
dx =
4x + 5
"25 − 16x2
=
dx
5 −1 4x
sin a b
4
5
5 −1 4x
1
sin a b − "25 − 16x2 + c
4
4
5
Exercise 8.6 Integrals involving inverse trigonometric functions
PRactise
1
WE27
a 3
Find:
1
"100 −
x2
dx
b 3
12
"64 −
9x2
dx.
2 Integrate each of the following with respect to x.
a
c08IntegralCalculus.indd 425
1
"36 − 25x2
b
x
"36 − 25x2
Topic 8 Integral calculus Integral calculus 425
14/08/15 4:51 PM
−2
On a certain curve the gradient is given by
. Find the equation
"25 − 16x2
of the curve that passes through the origin.
−2
4 On a certain curve the gradient is given by
. Find the equation of the
2
"4
−
x
curve that passes through the point (2, 3).
3
AOS 3
Topic 2
Concept 1
WE28
Antidifferentiate each of the following with respect to x.
1
12
a
b
2
100 + x
64 + 9x2
6 Integrate each of the following with respect to x.
5x
1
a
b
2
36 + 25x
64 + 25x2
7 WE30 Evaluate:
a 3
0
9
4
1
"81 − 16x2
b 3
dx
0
8 Evaluate:
9
WE31
E
#2
b 3
PA
G
"2
0
1
dx.
81 + 16x2
1
1
a 3
O
9
8
FS
WE29
PR
O
5
Antiderivatives
involving inverse
circular functions
Concept summary
Practice questions
1
dx
3 + 2x2
0
1
"2 − 3x2
dx.
Find each of the following with respect to x.
2
1
dx
25x − 20x + 13
2
EC
a 3
TE
D
1
1
dx
dx
b 3
4x − 12x + 25
"7 + 12x − 4x2
10 Find each of the following with respect to x.
a 3
R
R
11 WE32 Find 3
C
O
12 Find 3
U
N
Consolidate
5 − 3x
"25 − 9x2
dx.
1
"12 + 20x −
25x2
dx
3x + 5
dx.
9x2 + 25
13 Integrate each of the following with respect to x.
a
1
b
"16 − x2
1
"1 − 16x2
14 Find each of the following.
a 3
c 3
426 b 3
−1
"4 −
x2
dx
−3x
"36 −
49x2
dx
c
10
d
"49 − 25x2
b 3
d 3
−2
"1 −
4x2
10x
"49 − 25x2
dx
−3
"36 − 49x2
dx
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 426
14/08/15 6:20 PM
15 Find a set of antiderivatives for each of the following.
a
1
16 + x2
b
x
49 + 36x2
c
21
49 + 36x2
d
8
1 + 16x2
16 Evaluate each of the following.
3
a 3
0
3
x
"9 − x2
b 3
dx
0
3
1
"9 − x2
3
1
c 3
dx
9 + x2
dx
d 3
0
0
x
dx
9 + x2
1
. Find the equation of the
"4 − x2
curve that passes through the point ( !3, π) .
1
b On a certain curve the gradient is given by
. Find the equation of the
2
1
+
4x
curve that passes through the point a12, πb.
dy
1
c If
+
= 0, and when x = 1, y = 0, find y when x = 0.
dx 3 + x2
dy
1
d If
= 0, and when x = !3, y = 0, find y when x = 0.
+
dx "6 − x2
18 Evaluate each of the following.
0
5
c 3
1
"25 − 9x
2
3
x
"25 − 9x2
dx
EC
0
dx
TE
D
a 3
6
PA
G
5
E
PR
O
O
FS
17aOn a certain curve the gradient is given by
5
b 3
3
0
x
dx
25 + 9x2
5
d 3
3
0
1
dx
25 + 9x2
19 Evaluate each of the following.
1
R
a 3
R
0
1
"4 − 3x2
1
b 3
dx
0
1
1
dx
1 + 3x2
U
N
C
O
20 Find each of the following.
a 3
c 3
2
"5 − 4x −
x2
dx
6
"24 − 30x − 9x2
dx
21 Find each of the following.
a 3
c 3
3x − 4
"9 −
16x2
5 − 2x
"5 − 2x2
dx
dx
c 3
0
b 3
d 3
b 3
d 3
1
d 3
x
dx
1 + 3x2
x2
0
x
"4 − 3x2
dx
2
dx
+ 4x + 13
6
dx
74 + 30x + 9x2
3 + 4x
dx
9 + 16x2
5 − 2x
dx
25 + 2x2
Topic 8 Integral calculus Integral calculus 427
c08IntegralCalculus.indd 427
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22 If a, b, p and q are positive real constants, find each of the following.
a 3
c 3
"p − q x
2 2
ax + b
"p2 − q2x2
dx
b 3
dx
d 3
1
dx
p + q2x2
2
ax + b
dx
p + q2x2
2
23 a i Use the substitution x = 4 sin(θ) to find 3"16 − x2 dx.
iiEvaluate 3"16 − x2 dx. What area does this represent?
0
b i Use the substitution x =
6
5
6
5
cos(θ) to find 3"36 − 25x2 dx.
FS
4
O
Master
1
2
PR
O
iiEvaluate 3 "36 − 25x2 dx. What area does this represent?
0
c Prove that the total area inside the circle x2 + y2 = r2 is given by πr2.
2
x2 y
+
= 1 is given by πab.
a2 b2
PA
G
E
d Prove that the total area inside the ellipse
24 Evaluate each of the following.
5
3
x−4
dx
2
x − 6x + 13
4
1
dx
"8 + 2x − x2
b 3
3
2
2x + 1
dx.
4x − 12x + 25
2
1
d 3
−2
2x − 3
"12 − 8x − 4x2
dx
R
R
Integrals involving partial fractions
Integration of rational functions
O
8.7
3x + 2
EC
c 3
TE
D
a 3
7
2
C
U
N
AOS 3
Topic 2
Concept 7
Antiderivatives
with partial
fractions
Concept summary
Practice questions
428 A rational function is a ratio of two functions, both of which are polynomials. For
mx + k
is a rational function; it has a linear function in the numerator
example, 2
ax + bx + c
and a quadratic function in the denominator.
In the preceding section, we integrated certain expressions of this form when a > 0
and Δ = b2 − 4ac < 0, meaning that the quadratic function was expressed as the sum
of two squares. In this section, we examine cases when a ≠ 0 and Δ = b2 − 4ac > 0.
This means that the quadratic function in the denominator can now be factorised into
linear factors. Integrating expressions of this kind does not involve a new integration
technique, just an algebraic method of expressing the integrand into its partial
fractions decomposition.
Converting expressions into equivalent forms is useful for integration. We have seen
this when expressing a quadratic in the completing the square form or converting
trigonometric powers into multiple angles.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 428
14/08/15 6:21 PM
AOS 2
Topic 1
Concept 1
The method of partial fractions involves factorising the denominator, for example,
mx + k
mc + k
A
B
=
, and then expressing as
+
, where
2
(ax + α) (x + β)
ax + bx + c (ax + α)(x + β)
A and B are constants to be found. We can find A and B by equating coefficients and
then solving simultaneous equations, or by substituting in specific values of x. See the
following worked examples.
Equating coefficients means, for example, that if Ax + B = 3x − 4, then A = 3 from
equating the coefficient of x, since Ax = 3x, and B = −4 from the term independent
of x (the constant term).
33
Find:
a 3
b 3
12
dx
36 − x2
WRITE
PA
G
THINK
4
dx.
49 − 36x2
a 1 Factorise the denominator into linear factors
using the difference of two squares.
TE
D
2 Write the integrand in its partial fractions
decomposition, where A and B are constants
to be found.
EC
3 Add the fractions by forming a common
denominator.
R
4 Expand the numerator, factor in x and
O
R
expand the denominator.
C
5 Because the denominators are equal, the
U
N
numerators are also equal. Equate the
coefficients of x and the term independent
of x. This gives two simultaneous equations
for the two unknowns, A and B.
6 Solve the simultaneous equations
7 An alternative method can be used to
find the unknowns A and B. Equate the
numerators from the working above.
8 Substitute an appropriate value of x.
E
WORKED
EXAMPLE
PR
O
O
FS
Partial fractions
with a quadratic
denominator that
has linear factors
Concept summary
Practice questions
However, if the derivative of the denominator is equal to the numerator or
d
a constant multiple of it, that is, if (ax2 + bx + c) = 2ax + b, then
dx
2ax + b
dx = loge a ∣ax2 + bx + c∣b.
3 2
ax + bx + c
a
12
12
=
2
(6 + x)(6 − x)
36 − x
=
A
B
+
6+x 6−x
=
A(6 − x) + B(6 + x)
(6 + x)(6 − x)
6A − 6Ax + 6B + 6Bx
(6 + x)(6 − x)
6x(B − A) + 6(A + B)
=
36 − x2
=
6x(B − A) + 6(A + B) = 12
(1) B − A = 0
(2) 6(A + B) = 12
(1) ⇒ A = B; substitute into
(2) 12A = 12 ⇒ A = B = 1
12 = A(6 − x) + B(x + 6)
Substitute x = 6:
12 = 12B
B =1
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 429
429
14/08/15 6:22 PM
Substitute x = −6:
12 = 12A
A=1
12
1
1
=
+
2
6+x 6−x
36 − x
9 Substitute an appropriate value of x.
10 Express the integrand as its partial
fractions decomposition.
12
1
1
336 − x2 dx = 3 a 6 + x + 6 − x b dx
11 Instead of integrating the original expression,
we integrate the partial fractions expression,
since these expressions are equal.
= loge a ∣6 + x∣b − loge a∣6 − x∣b + c
O
1
1
dx = loge a ∣ax + b∣b.
a
ax + b
∣ ∣
13 Use the log laws to express the final
answer as a single log term.
b 1 Factorise the denominator into linear
b
4
4
=
2
(7 − 6x)(7 + 6x)
49 − 36x
fractions decomposition, where A and B
are constants to be found.
3 Add the fractions by forming a common
=
TE
D
denominator.
=
PA
G
2 Write the integrand in its partial
EC
5 Because the denominators are equal, the
7A + 6Ax + 7B − 6Bx
(7 − 6x)(7 + 6x)
6x(A − B) + 7(A + B)
=
49 − 36x2
6x(A − B) + 7(A − B) = 4
(1) A − B = 0
(2) 7(A + B) = 4
O
R
R
numerators are also equal. Equate coefficients
of x and the term independent of x. This
gives two simultaneous equations for the two
unknowns, A and B.
A(7 + 6x) + B(7 − 6x)
(7 − 6x)(7 + 6x)
=
4 Expand the numerator, factor in x and expand
the denominator.
A
B
+
7 − 6x 7 + 6x
E
factors using the difference of two squares.
6+x
12
336 − x2 dx = loge a 6 − x b + c
PR
O
the result 3
FS
12 Integrate term by term, using
U
N
C
6 Solve the simultaneous equations
7 Express the integrand as its partial
fractions decomposition.
8 Instead of integrating the original expression,
we integrate the partial fractions expression,
since these expressions are equal. Transfer
the constant factors outside the integral sign.
(1) ⇒ A = B: substitute into
(2) 14A = 4 ⇒ A = B = 27
4
2
2
=
+
2
7(7
−
6x)
7(7
+
6x)
49 − 36x
4
349 − 36x2 dx
= 3a
2
2
+
bdx
7(7 − 6x) 7(7 + 6x)
2
1
1
= 3a
+
bdx
7 7 − 6x 7 + 6x
430 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 430
14/08/15 4:43 PM
= 27 c −16 loge a ∣7 − 6x∣b + 16 loge a ∣7 + 6x∣b d + c
9 Integrate term by term, using
the result 3
1
1
dx = loge a ∣ax + b∣b.
a
ax + b
7 + 6x
1
4
349 − 36x2 dx = 21 loge a ` 7 − 6x ` b + c
10 Use the log laws to express the final answer
as a single log term.
34
Find 3
x + 10
dx.
x2 − x − 12
WRITE
1 Factorise the denominator into linear factors.
PR
O
THINK
O
WORKED
EXAMPLE
FS
Compare the previous worked example to Worked examples 27 and 29.
x + 10
x + 10
=
x − x − 12 (x − 4)(x + 3)
2
=
3 Add the fractions by forming a common
denominator.
expand the denominator.
TE
D
4 Expand the numerator, factor in x and
PA
G
decomposition, where A and B are constants
to be found.
5 Because the denominators are equal,
=
A
B
+
x−4 x+3
A(x + 3) + B(x − 4)
(x − 4)(x + 3)
Ax + 3A + Bx − 4B
(x − 4)(x + 3)
x(A + B) + 3A − 4B
=
x2 − x − 12
x(A + B) + 3A − 4B = x + 10
(1) A + B = 1
(2) 3A − 4B = 10
=
U
N
C
O
R
R
EC
the numerators are also equal.
Equate the coefficients of x and the
term independent of x to give two
simultaneous equations for the two
unknowns, A and B.
6 Solve the simultaneous equations by
elimination. Add the two equations to
eliminate B and solve for A.
E
2 Write the integrand in its partial fractions
7 Substitute back into (1) to find B.
8 An alternative method can be used to
find the unknowns A and B. Equate the
numerators from the working above.
4 × (1) 4A + 4B = 4
(2) 3A − 4B = 10
4 × (1) + (2) 7A = 14
A= 2
2+B= 1
B = −1
A(x + 3) + B(x − 4) = x + 10
9 Let x = 4.
7A = 14
A= 2
10 Let x = −3.
−7B = 7
B = −1
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 431
431
14/08/15 6:21 PM
x + 10
2
1
3x2 − x − 12 dx = 3 a x − 4 − x + 3 bdx
11 Express the integrand as its partial
fractions decomposition.
x + 10
3x2 − x − 12 dx
12 Integrate term by term using the result
1
1
3ax + b dx = a loge a ∣ax + b∣b, and state
the final simplified answer using log laws.
(x − 4) 2
b
∣x + 3 ∣
mx + k
when a ≠ 0 and Δ = b2 − 4ac = 0, the
ax + bx + c
quadratic function in the denominator is now a perfect square.
2
O
AOS 2
Topic 1
PR
O
Concept 2
Integrating by the method of partial fractions in this case involves writing
mx + k
mc + k
A
B
=
and then expressing it as
+
, where
2
2
( px + α) ( px + α) 2
ax + bx + c ( px + α)
Partial
fractions with
perfect square
denominators
Concept summary
Practice questions
E
A and B are constants to be found. We can find A and B by equating coefficients and
then solving the simultaneous equations, as before.
35
Find 3
6x − 5
dx.
4x − 12x + 9
2
TE
D
THINK
PA
G
WORKED
EXAMPLE
= loge a
1 Factorise the denominator as a perfect
square.
WRITE
4x2
6x − 5
6x − 5
=
− 12x + 9 (2x − 3) 2
=
EC
2 Write the integrand in its partial fractions
decomposition, where A and B are constants
to be found.
R
R
=
C
O
denominator and expanding the numerator.
U
N
4 Because the denominators are equal, the numerators
are also equal. Equate coefficients of x and the term
independent of x to give two simultaneous equations
for the two unknowns, A and B.
5 Solve the simultaneous equations.
6 Express the integrand as its partial fractions
decomposition.
A
B
+
2x − 3 (2x − 3) 2
A(2x − 3) + B
(2x − 3) 2
2Ax + B − 3A
=
(2x − 3) 2
3 Add the fractions by forming the lowest common
432
FS
Perfect squares
If an expression is of the form
= 2 loge a ∣x − 4∣b − loge a ∣x + 3∣b + c
2Ax + B − 3A = 6x − 5
(1) 2A = 6
(2) B − 3A = −5
(1) ⇒ A = 3: substitute into (2):
B − 9 = −5 ⇒ B = 4
6x − 5
34x2 − 12x + 9 dx
= 3a
3
4
+
bdx
2x − 3 (2x − 3) 2
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 432
14/08/15 6:21 PM
7 Integrate term by term, using the result
1
1
3ax + b dx = a loge a ∣ax + b∣ b in the first term and
3
=
3
2
loge a ∣2x − 3∣b −
+c
2
2x − 3
1
1
for the second term.
dx = −
2
a(ax + b)
(ax + b)
Note that this example can also be done using a linear substitution with a back
substitution. See Worked example 12b.
36
Find 3
2x2 + 5x + 3
dx.
x2 + 3x − 4
THINK
TE
D
WORKED
EXAMPLE
PA
G
E
PR
O
O
FS
Rational functions involving ratios of two quadratic functions
When the degree of the polynomial in the numerator is greater than or equal to the
degree of the polynomial in the denominator, the rational function is said to be a
non-proper rational function. In this case we have to divide the denominator into the
numerator to obtain a proper rational function.
For example, when we have quadratic functions in both the numerator and
rx2 + sx + t
denominator, that is, the form 2
where r ≠ 0 and a ≠ 0, we can use long
ax + bx + c
division to divide the denominator into the numerator to express the function as
rx2 + sx + t
mx + k
r
=q+ 2
where q = .
2
a
ax + bx + c
ax + bx + c
1 Use long division to express the rational function
R
R
EC
as a proper rational function.
C
O
2 Factorise the denominator into linear factors.
U
N
3 Write the integrand in its partial fractions
decomposition, where A and B are constants
to be found.
WRITE
2x2 + 5x + 3
x2 + 3x − 4
2(x2 + 3x − 4) + 11 − x
=
x2 + 3x − 4
11 − x
=2+ 2
x + 3x − 4
11 − x
=2+
(x − 1)(x + 4)
=2+
A
B
+
x−1 x+4
A(x + 4) + B(x − 1)
(x − 1)(x + 4)
common denominator and expanding the
x(A + B) + 4A − B
numerator.
=2+
x2 + 3x − 4
5 Because the denominators are equal, the numerators x(A + B) + 4A − B = 11 − x
are also equal. Equate coefficients of x and the
(1) A + B = −1
term independent of x to give two simultaneous
(2) 4A − B = 11
equations for the two unknowns, A and B.
4 Add the fractions by forming a
=2+
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 433
433
14/08/15 6:21 PM
5A = 10
A = 2 ⇒ B = −3
6 Solve the simultaneous equations by elimination.
Add the two equations to eliminate B.
2x2 + 5x + 3
3 x2 + 3x − 4 dx
7 Express the integrand as its partial
fractions decomposition.
= 3 a2 +
2x2 + 5x + 3
3 x2 + 3x − 4 dx
8 Integrate term by term using the result
1
1
3ax + b dx = a loge a ∣ax + b∣ b and state
the final answer.
FS
O
(x − 1) 2
b+c
(x + 4) 3
PR
O
= 2x + loge a
E
Concept 3
PA
G
Partial fractions
with quadratic
factors in the
denominator
Concept summary
Practice questions
37
Find 3
TE
D
equations to solve for the three unknowns: A, B and C.
x2 + 7x + 2
dx.
x3 + 2x2 + 4x + 8
EC
THINK
1 First try to factorise the cubic in the
O
R
R
denominator using the factor theorem.
U
N
C
2 Factorise the denominator into factors.
3 Write the integrand in its partial fractions
decomposition, where A, B and C are
constants to be found.
4 Add the fractions by forming a common
denominator and expanding the numerator.
434
= 2x + 2 loge a ∣x − 1∣b − 3 loge a ∣x + 4∣b + c
Rational functions involving non-linear factors
If the denominator does not factorise into linear factors, then we proceed with
px2 + qx + r
Bx + C
A
=
+
. Note that we will need three simultaneous
(x + α)(x2 + a2) x + α x2 + a2
AOS 2
Topic 1
WORKED
EXAMPLE
3
2
−
b dx
x−1 x+4
WRITE
f(x) = x3 + 2x2 + 4x + 8
f(1) = 1 + 2 + 4 + 8 ≠ 0
f(2) = 8 + 8 + 8 + 8 ≠ 0
f(−2) = −8 + 8 − 8 + 8 = 0
(x + 2) is a factor.
x3 + 2x2 + 4x + 8 = (x + 2)(x2 + 4)
x2 + 7x + 2
x3 + 2x2 + 4x + 8
Bx + C
A
=
+ 2
x+2
x +4
A(x2 + 4) + (x + 2)(Bx + C)
(x + 2)(x2 + 4)
Ax2 + 4A + Bx2 + 2Bx + Cx + 2C
=
(x + 2)(x2 + 4)
x2 (A + B) + x(2B + C) + 4A + 2C
=
x3 + 2x2 + 4x + 8
=
MATHS QUEST 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 434
14/08/15 6:21 PM
(1) A + B = 1
the numerators are also equal. Equate
(2) 2B + C = 7
the coefficients of x2 and x and the term
(3) 4A + 2C = 2
independent of x to give three simultaneous
equations for the three unknowns: A, B and C.
5 Because the denominators are equal,
2 × (1) 2A + 2B = 2
(2) 2B + C = 7
Subtracting gives
(4) C − 2A = 5
6 Solve the simultaneous equations by
elimination.
Substitute into (2):
2B + 3 = 7 ⇒ B = 2
Substitute into (1): ⇒ A = −1
remaining unknowns.
decomposition.
EC
TE
D
9 Separate the last term into two expressions.
10 Integrate term by term, using the
dx = loge a∣ f(x) ∣b,
R
f ′(x)
x2 + 7x + 2
3x3 + 2x2 + 4x + 8 dx
= 3a
= 3a
2x + 3
−1
+ 2
bdx
x+2 x +4
x2
3
2x
1
−
+ 2
bdx
+4 x+2 x +4
x2 + 7x + 2
dx
3 3
x + 2x2 + 4x + 8
= loge (x2 + 4) − loge a ∣x + 2∣b +
3
x
tan −1 a b + c
2
2
U
N
C
O
R
f(x)
1
1
3ax + b dx = a loge a ∣ax + b∣b and
x
1
1
dx = tan−1 a b, and state the
3 2
2
a
a
x +a
final answer.
x2 + 7x + 2
2x + 3
−1
=
+ 2
3
2
x + 2x + 4x + 8 x + 2 x + 4
PA
G
8 Express the integrand as its partial fractions
E
7 Back substitute to find the
results 3
O
PR
O
(3) 4A + 2C = 2
Adding gives 4C = 12
C=2
FS
2 × (4) 2C − 4A = 10
Exercise 8.7 Integrals involving partial fractions
PRactise
1
WE33
a 3
Find:
1
dx
100 − x2
2 Find:
a 3
1
dx
36 − 25x2
b 3
12
dx.
64 − 9x2
b 3
x
dx.
36 − 25x2
Topic 8 Integral calculus Integral calculus 435
c08IntegralCalculus.indd 435
14/08/15 6:22 PM
Find 3
Find 3
6 Find 3
7
WE36
9
WE37
3x2 + 10x − 4
dx.
x2 + 3x − 10
−2x2 − x + 20
dx.
x2 + x − 6
Find 3
10 Find 3
+
x3
19 − 3x
dx.
x − 2x2 + 9x − 18
3
25
dx.
+ 16x + 48
3x2
11 Integrate each of the following with respect to x.
a
x2
x + 11
+ x − 12
b
x2
5x − 9
− 2x − 15
TE
D
Consolidate
2x − 1
dx.
9x − 24x + 16
2
Find 3
8 Find 3
2x + 1
dx.
x + 6x + 9
2
FS
WE35
O
5
x − 11
dx.
x + 3x − 4
2
PR
O
4 Find 3
x + 13
dx.
x + 2x − 15
2
E
WE34
PA
G
3
c
2x − 19
+x−6
d
x2
x2
11
− 3x − 28
12 Find an antiderivative of each of the following.
a
x2
1
−4
b
2
16 + x2
c
x2
x
− 25
d
2x − 3
x2 − 36
EC
13 Integrate each of the following with respect to x.
2x + 3
− 6x + 9
4x
c
2
4x + 12x + 9
2x − 5
+ 4x + 4
6x − 19
d
2
9x − 30x + 25
b
R
x2
O
R
a
x2
U
N
C
14 Find an antiderivative of each of the following.
a
x2 − 4x − 11
x2 + x − 12
b
−3x2 − 4x − 5
x2 + 2x − 3
c
4x2 − 17x − 26
x2 − 4x − 12
d
−2x3 + 12x2 − 17x
x2 − 6x + 8
15a Determine 3
ik = 0
b Determine 3
x2
1
dx for the cases when:
+ kx + 25
iik = 26
1
dx for the cases when:
4x − 12x + k
2
ik = 8
436 iiik = −10.
iik = 9
iiik = 25.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 436
14/08/15 4:44 PM
16 Evaluate each of the following.
2
a 3
1
6
b 3
1
dx
x + 4x
2
5
1
1
dx
x − 16
2
4
3x + 8
c 3
dx
x2 + 6x + 8
d 3
−1
3
x−6
dx
x2 − 4x + 4
5
, the coordinate
+
x
−
6
axes and x = −1.
2x − 3
b Find the area bounded by the curve y =
, the x-axis and the
2
4
+
3x
−
x
lines x = 2 and x = 3.
21
c Find the area bounded by the curve y =
, the coordinate
2
40
−
11x
−
2x
axes and x = −5.
x3 − 9x + 9
d Find the area bounded by the curve y =
, the x-axis and the
2
x
−
9
lines x = 4 and x = 6.
x2
PR
O
O
FS
17aFind the area bounded by the curve y =
1
2
b Find the value of b if 3
2
dx = loge (a).
4x − x2
PA
G
18 a Find the value of a if 3
2
"4x −
TE
D
1
4
EC
c Find the value of c if 3
3
x2
R
R
O
C
U
N
i3y dx
b If y =
i3y dx
x2
dx = πb.
3x
dx = loge (c) .
−x−2
2
d Find the value of d if 3
19 a If y =
E
2
1
x2
3
dx = πd.
−x+1
2
+ 25
, find:
16x2 − 25
16x2
ii3 dx.
1
y
36x2 − 49
, find:
36x2 + 49
ii3 dx.
1
y
1
20 Show that 3
0
x4 (1 − x) 4
22
dx =
− π.
2
7
1+x
Topic 8 Integral calculus Integral calculus 437
c08IntegralCalculus.indd 437
14/08/15 4:44 PM
Master
21 Evaluate each of the following.
1
a 3
0
2
b 3
2
27x
dx
16 − 9x2
0
3
20x2
dx
4x2 + 4x + 1
3
x2 − 2x + 9
c 3
dx
x3 + 9x
d 3
!3
2
4x2 − 16x + 19
dx
(2x − 3) 3
22 If a, b, p and q are all non-zero real constants, find each of the following.
FS
O
d 3
x
dx
(ax − b) 2
x
dx
b x − a2
2 2
1
dx
( px + a)(qx + b)
U
N
C
O
R
R
EC
TE
D
PA
G
E
c 3
b 3
1
dx
b x − a2
2 2
PR
O
a 3
438 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 438
14/08/15 4:44 PM
ONLINE ONLY
8.8 Review
The Maths Quest Review is available in a customisable
format for you to demonstrate your knowledge of this
topic.
The Review contains:
• Short-answer questions — providing you with the
opportunity to demonstrate the skills you have
developed to efficiently answer questions using the
most appropriate methods
• Multiple-choice questions — providing you with the
opportunity to practise answering questions using
CAS technology
www.jacplus.com.au
• Extended-response questions — providing you with
the opportunity to practise exam-style questions.
A summary of the key points covered in this topic is
also available as a digital document.
FS
REVIEW QUESTIONS
Units 3 & 4
Integral calculus
Sit topic test
U
N
C
O
R
R
EC
TE
D
studyON is an interactive and highly visual online
tool that helps you to clearly identify strengths
and weaknesses prior to your exams. You can then
confidently target areas of greatest need, enabling
you to achieve your best results.
PA
G
E
PR
O
O
Download the Review questions document from
the links found in the Resources section of your
eBookPLUS.
Topic 8 INTEGRAL CALCULUS
c08IntegralCalculus.indd 439
439
14/08/15 4:45 PM
8 Answers
1
3 y = 8 (!16x + 25 − 5)
Exercise 8.2
11
1 26
4 4
16
2 3
92
3 3
5 3 loge a ∣3x − 5∣b + c
1
4 16
6 −2 loge a ∣7 − 2x∣b + c
5 4
8 8
1
1
− 1)
− e−2)
b
9 6
10
1372
27
12128
114
1
131212
12
1436
16a i 32
16623
18a i loge (4)
1
1
b 813
c i648
21a
4a3
3
1
6
b
1 4
a
2
b
1
2
d
2
3
O
1
ii 1212
− 2 loge (4) ≈ 0.9774
U
N
− 13 loge (9) ≈ 0.7491
24Check with your teacher.
25a i (3.4443, 3.9543)
b
3
10
b i (3.8343, 1.9172)
26a i (2.6420, 86.1855)
−1
+c
6(3x + 5) 2
−1
+c
3(3x + 5)
d
1
(6x
54
+ 7) 9 + c
b
loge a ∣6x + 7∣b + c
d
loge a ∣3x + 4∣b + c
16a
c
1
(3x
21
1
6
1 3
" (3x
2
1
!6x
3
+ 5) 2 + c
+7+c
−1
+c
6(6x + 7)
1
5
(3x + 5) 8 − 63
(3x + 5) 7 + c
72
1
= 504 (21x − 5)(3x + 5) 7 + c
5
1
loge a ∣3x + 5∣b +
+c
9
9(3x + 5)
−(6x + 5)
c
+c
18(3x + 5) 2
b
2
d
18a
ii 4.6662
ii 8.6558
ii 64.8779
b i(7.5882, –1.7003), (24.2955, –4.6781)
ii 384.3732
Exercise 8.3
1
1 35 (5x − 9) 7 + c
1
2 24 (3x + 4) 8 + c
440 c
17a
ii 288
12112
15
4
40
27
3
ii 465 4
iii
85 13
C
23a
iii loge (a)
c
22a i288
b i
2a
n
1
ii 1 −
a
R
19a 1392
20a
c
ii 1
2
3
b i
15a
TE
D
16
π
b
b
8
3
− 5) !6x + 5 + c
3
ii 132
36
π
+ 5) 7 + c
loge a 25 b − 1
142 + 5 loge a 5 b
EC
17a
ii 22
R
b i
1
(3x
27
+c
PA
G
b 37 2
b
8(2x + 7) 2
132x −
1
15a 9
5
3
3
+ 50
(5x − 9) 6 + c
3)(5x − 9) 6
5
2
loge a ∣3x − 4∣b −
+c
9
9(3x − 4)
−(4x + 7)
10a
118
1
(5x − 9) 7
175
1
= 350
(10x +
O
8 6(e2
9a PR
O
7 2(e6
E
6 12
FS
7 2
1
(2x
10
− 5)(3x + 5) 3 + c
1
(6x
360
7
+ 7) 10 − 324
(6x + 7) 9 + c
1
= 3240
(54x − 7)(6x + 7) 9 + c
b
1
(3x
27
− 7) !6x + 7 + c
x
7
−
loge a ∣6x + 7∣b + c
6 36
7
1
d
+
loge a ∣6x + 7∣b + c
36(6x + 7) 36
t
b y = −!3 − 2x
19a
2(2 − 5t)
c
c
3
2
loge (3)
1
d y = 3 (x − 9) !2x + 9 + 9
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 440
14/08/15 4:46 PM
b
c
32
3
256
15
d i
ii
c
d
25a
b
c
d
b 2 sin(!x) + c
+c
b
1
( loge
8
(3x)) 2 + c
π2
24
π2
12
128
11
3
3
2
(3ax − 3b)(ax + b) 2 + c
15a2
1
loge a ∣ax + b∣b + c
a
x
b
− loge a ∣ax + b∣b + c
a a2
2
!ax + b + c
a
2
(ax − 2b) !ax + b + c
3a2
−1
+c
a(ax + b)
b
1
+ loge a ∣ax + b∣b + c
2
a (ax + b) a2
ad − bc
cx
loge a∣ax + b∣b +
+k
a
a2
c
ad − bc
loge a ∣ax + b∣b −
+k
2
2
a
a (ax + b)
a2d + cb2
cx 2bc
−
loge a ∣ax + b∣b −
+k
a2
a3
a3 (ax + b)
a2d + cb2
cx2 bcx
loge a ∣ax + b∣b +
−
+k
2a
a3
a2
b2
1
(ax − 3b)(ax + b) + loge a ∣ax + b∣b + c
2a3
a3
x(xa + 2b) 2b
b
−
loge a ∣ax + b∣b + c
a2 (ax + b)
a3
b(4ax + 3b)
1
c
loge a ∣ax + b∣b +
+c
3
a
2a3 (ax + b) 2
2
d
(3a2x2 − 4abx + 8b2) !ax + b + c
15a3
O
C
U
N
Exercise 8.4
−2
+c
(x2 + 16) 2
5
2 2"2x2 + 3 + c
1
3 2 loge (x2 + 4x + 29) + c
1
4 3
+c
1080
26a
1
1 sin(3x)
e
3
1 tan(2x)
e
2
1
4
"a3
3
8
"a5
15
loge a ∣
13a
x3
+ 9∣b + c
1 2
(x
12
+ 4) 6 + c
b
−1
+c
+ 9)
−1
14a
+c
6(x3 + 27) 2
c
c
15a
TE
D
b
1
6 2 sin(x2) + c
9 1
EC
24a
d 6
R
d
1
x
5 cosa b + c
8a
R
c
1
35
333
10
7a
2
23a
(ax + b) 2 + c
3a
b
c
d
2(x2
1
3
FS
ii
c 6
loge a ∣ x3 + 8∣b + c
1 2
(x
8
1
2
loge (x2 + 4) + c
O
i
b
16
3
32
3
47
7
1
loge a 14
b
3
5
PR
O
22a
b
− 4x + 13) 4 + c
E
21a
7
2
26
3
d "x2 + 9 + c
b
d
b
2
"x3
3
+ 27 + c
1 3
(x
12
+ 8) 4 + c
2(x2
−1
+c
− 4x + 13)
c −"x2 − 8x + 25 + c
PA
G
20a
1
d −2 loge (x2 − 8x + 25) + c
16a
c
1
loge
8
(4e2x + 5) + c
1
+c
12(3e−2x + 4) 2
17a −cos(loge (4x)) + c
b
d
x
1
asin−1 a b b + c
2
2
1
18a y = 2 (1 − cos(x2))
c
1
2
loge a 29
b
20
19a12
20a
1
2
b loge a 26
b
29
loge a 53 b c 2e(e − 1)
21a
c
22a
c
1
(e
2
1
(1
2
d
+c
−1
+c
e2x + x
1
(tan−1 (2x)) 2
4
+c
b 1 − !3
d
1
4
− 5)
b sin(loge (3x)) + c
2
c
1
6(2e−3x
1
(1
2
c − e)
1
18
d 0
1
b −4
d
1
(e
2
1
6
− 1)
− 1)
b
− !3)
d 2(cos(1) − cos(2))
c = 4, 12 loge a 21
b
5
1
(1 − e−4)
2
b
!2
4
d 2
−1
1
b
+c
"ax2 + b + c
2a(ax2 + b)
a
1
1
(ax2 + b) n+1 + c d
loge a ∣ax2 + b∣b + c
c
2a(n + 1)
2a
23a
24a loge a ∣ f(x) ∣ b + c
c 2!f(x) + c
b −
1
+c
f(x)
d e f(x) + c
Topic 8 Integral calculus Integral calculus calculus Integral calculus 441
c08IntegralCalculus.indd 441
14/08/15 5:20 PM
c
1
20
4π − 3!3
12
b
1
3 24 sin6 (4x) + c
!3
π
−
24 128
21a
b
1
4 8
1
5 12 sin3 (4x) −
1
sin5 (4x)
10
c
1
+ 28
sin7 (4x) + c
22a
8 − 5!2
36
3x 1
1
7
− sin(4x) +
sin(8x) + c
8
8
64
π
1
8
+
32 12
6
b
c
d
x
2
9 2 loge a ` seca b ` b + c
23a
1
104 loge (2)
b
x
2
113 tana b − x + c
d
1
13a −8 cos(4x) + c
c
14a
−18 cos4 (2x)
1
10
b x + c
+c
b
d
π
c
32
1
15
c
π
12
π
c
16
1
18a x − 4 cos(4x) + c
b
U
N
C
17a
1
(sin(2x)
2
c
442 b
d
1
sec(2x) + c
2
−14 cosec2 (2x)
b
2
9
d
8
45
+c
1
1
cos(4x)
32
+ c1
ii
1
cos6 (2x)
12
1
cos4 (2x)
8
+ c3
iii
1
8
−
∣sec(3x) ∣b + c
1
tan2 (3x)
6
+c
b
d
1
loge a
3
b
c
d
1a sin−1 a
2a
x
b+c
10
1 −1 5x
sin a b + c
5
6
x
2
∣sin(3x) ∣b + c
1
tan3 (3x)
9
8
75
Exercise 8.6
1
2
1
sin4 (2x) − 12
sin6 (2x) + c3
1
loge a
3
d
−1
cosn+1 (ax) + c
a(n + 1)
1
sinn+1 (ax) + c
a(n + 1)
1
tann+1 (ax) + c
a(n + 1)
1
loge a ∣sec(ax) ∣ b + c
a
1
tan(ax) − x + c
a
1
1
tan2 (ax) + loge a ∣cos(ax) ∣b + c
a
2a
1
1
tan3 (ax) − tan(ax) + x + c
a
3a
x
1
−
sin(2ax) + c
2 4a
1
1
cos3 (ax) − cos(ax) + c
a
3a
3x
1
1
−
sin(2ax) +
sin(4ax) + c
8
4a
32a
1
2
1
− cos(ax) +
cos3 (ax) −
cos5 (ax) + c
a
3a
5a
x
1
+
sin(2ax) + c
2 4a
1
1
sin(ax) −
sin3 (ax) + c
a
3a
3x
1
1
+
sin(2ax) +
sin(4ax) + c
8
4a
32a
1
2
1
sin(ax) −
sin3 (ax) +
sin5 (ax) + c
a
3a
5a
3 − sin−1 a
− cos(2x)) + 16 (cos3 (2x) − sin3 (2x)) + c
c i 96 cos3 (4x) −
19a
d
1
15
R
+c
R
16a
+c
O
d
1
1
sin3 (4x) − 20
sin5 (4x)
12
1
1
sin5 (4x) − 28
sin7 (4x)
20
−12 cosec(2x) + c
1
sec2 (2x) + c
4
+c
EC
x
1
15a
−
sin(16x) + c
8 128
1
1
b 20 cos5 (4x) − 12 cos3 (4x) + c
c
24a
1
sin4 (2x)
8
4−π
20
PA
G
4−π
16
TE
D
12
c
b
FS
x
2
b −cosa b + c
loge (2)
O
2a
x
x
− sina b + c
2
2
1
10
PR
O
1a
20a
E
Exercise 8.5
+c
1
3x
b+c
8
b −25"36 − 25x2 + c
4x
4x
1
b = a2 cos−1 a b − πb
5
4
5
4 2 cos−1 a b + 3
x
1
tan−1 a b + c
10
10
5x
1
6a
tan−1 a b + c
30
6
π
7a
24
π !6
8a
36
5a
9a
b 4 sin−1 a
1 −1 2x − 3
tan a
b+c
8
4
b
b
1 −1 3x
tan a b + c
2
8
1
loge
10
(64 + 25x2) + c
π
144
π !3
b
9
b
b
1 −1 2x − 3
sin a
b+c
2
4
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 442
14/08/15 6:22 PM
x
2
b
5x
b+c
7
3
"36
49
2
b cos−1 (2x) + c
− 49x2 + c
d
1 −1 x
tan a b + c
4
4
1 −1 6x
c
tan a b + c
2
7
15a
b
sin−1
π
2
d
1
2
3x + 5
b+c
7
1
18
d
π
60
b
!3 π
9
loge (2)
1
3
x+2
2
b tan−1 a
b+c
3
3
d
2 −1 3x + 5
tan a
b+c
7
7
C
U
N
b
c
d
22a
c
x + 10
1
loge a `
`b + c
20
x − 10
3x + 8
1
loge a `
`b + c
4
3x − 8
5x + 6
1
loge a `
`b + c
60
5x − 6
b −50 loge a ∣25x2 − 36∣b + c
1
3
4x
b − "9 − 16x2 + c
3
16
1 −1 4x
1
tan a b + loge (16x2 + 9) + c
4
3
8
5!2 −1 !10 x
"5 − 2x2 +
sin a
b+c
2
5
!2 −1 !2 x
1
tan a
b − loge (2x2 + 25) + c
2
5
2
qx
1 −1 qx
1
sin a b + c
b
tan−1 a b + c
q
p
pq
p
a
b −1 qx
sin a b − "p2 − q2x2 + c
q
p
q2
21a cos−1 a
2a
loge (2)
b
d
x+2
a
b+c
3
c 2 sin−1 a
b
R
20a 2
(2)
R
1
loge
3
1a
b
EC
5
9
O
c
(36x2 + 49) + c
7π
1
b y = tan−1 (2x) +
2
8
π
d
4
2π
x
17a y = sin−1 a b +
2
3
π !3
c
18
π
18a
18
!3 π
9
1
loge
72
Exercise 8.7
PA
G
π
c
12
19a
7x
3
cos−1 a b + c
7
6
d 2 tan−1 (4x) + c
16a 3
c
sin−1 (4x) + c
d −5 "49 − 25x2 + c
14a cos−1 a b + c
c
1
4
FS
x
4
13a sin−1 a b + c
c 2 sin−1 a
5x
x
18
cos−1 a b + c
2
5
6
9π
ii ; one-quarter of the area of an ellipse with
5
semi-minor and major axes of 65 and 6
π
π
1
1
24a
b loge (2) +
loge (2) −
2
8
4
8
5π
5π
c 9 +
d !3 −
2
3
b i "36 − 25x2 −
1 −1 3x
tan a b + c
3
5
12 loge (9x2 + 25) +
x
x
4
2
ii 4π; one-quarter of the area of a circle of radius 4
23a i 8 sin−1 a b + "16 − x2 + c
O
1
6
3x
1
b + "25 − 9x2 + c
5
3
1 −1 5x − 2
sin a
b+c
5
3
PR
O
11 sin−1 a
b
E
5
3
5x − 2
1
tan−1 a
b+c
15
3
TE
D
10a
qx
a
b
tan−1 a b +
d
loge (q2x2 + p2) + c
pq
p
2q2
3 loge a
4 loge a
(x − 3) 2
∣x + 5∣
∣x + 4∣3
(x − 1) 2
b+c
b+c
5
+ 2 loge a ∣x + 3∣b + c
x+3
5
2
6 loge a ∣3x − 4∣ b −
+c
9
9(3x − 4)
5
7 3x + loge a
8 loge a
9 loge a
10loge a
(x − 2) 4
∣x + 5∣3
(x − 2) 2
∣x + 3∣
∣x − 2∣
b − 2x + c
"x2 + 9
∣x + 3∣
b − 5 tan−1 a x b + c
"x2 + 16
11a loge a
b+c
3
3
b + 3 tan−1 a x b + c
(x − 3) 2
∣x + 4∣
4
4
b+c
b loge a (x − 5) 2 ∣x + 3∣3b + c
c loge a
d loge a
∣x + 3∣5
b+c
∣x − 2∣3
∣ ∣
x−7
b+c
x+4
Topic 8 Integral calculus Integral calculus calculus Integral calculus 443
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∣ ∣
x−2
1
loge a
b+c
4
x+2
16a
x
1
b tan−1 a b + c
2
4
d
1
4
loge a ∣ (x − 6) 3 (x + 6) 5∣ b + c
9
+c
x−3
9
b 2 loge a ∣x + 2∣ b +
+c
x+2
3
c loge a ∣2x + 3∣b +
+c
2x + 3
3
2
d loge a ∣3x − 5∣b +
+c
3
3x − 5
∣x + 2∣3
d −x2 + loge a
1
4
1 −1 2x − 3
tan a
b+c
8
4
O
π
1
loge (3) −
2
18
b 12 − 5 loge (5)
d
2
9
+ 12 loge (3)
d
∣qx + b∣
1
loge a
b+c
aq − bp
∣ px + a∣
U
N
C
O
iii
1
+c
2(3 − 2x)
2 !3
3
∣bx − a∣
1
loge a
b+c
2ab
∣bx + a∣
1
b
loge a ∣b2x2 − a2∣ b + c
2b2
b
1
c
loge a ∣ax − b∣b −
+c
2
2
a
a (ax − b)
22a
b+c
x−2
`b + c
x−1
1
3
R
ii
c
1
+c
x−5
b i loge a `
d
21a 2 loge (7) − 3
`b + c
x+1
1
loge a `
`b + c
24
x + 25
iii −
c 5
b
20Check with your teacher.
TE
D
ii
1 −1 x
tan a b + c
5
5
7
3
EC
i d 10 + 2 loge a 3 b
3
R
15a (x − 4) 2
c loge (8)
b loge a 2 b
5
4x − 5
loge a `
`b + c
4
4x + 5
5
4x
ii x − tan−1 a b + c
2
5
7 −1 6x
b i x − tan a b + c
3
7
6x − 7
7
ii x + loge a `
`b + c
6
6x + 7
b+c
∣x − 2∣
9
PA
G
c 4x + loge a
(x − 1) 3
(x − 6) 2
d loge (2) − 2
19a i x +
14a x − loge a ∣ (x − 3) 2 (x + 4) 3∣b + c
(x + 3) 5
loge a 95 b
25
18a !3
13a 2 loge a ∣x − 3∣ b −
1
8
c loge a 3 b
17a loge a 4 b
loge a ∣x2 − 25∣ b + c
b −3x + loge a `
b
FS
1
2
loge a 53 b
PR
O
c
1
4
E
12a
444 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c08IntegralCalculus.indd 444
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FS
O
PR
O
E
PA
G
TE
D
EC
R
R
O
C
U
N
c08IntegralCalculus.indd 445
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