Block 8: Advanced Integration
Contents
8.1
8.2
8.3
8.4
8.5
Revision of
Integration
Integration
Integration
Answers to
Integration . . . . . . .
by Substitution . . . . .
by Parts . . . . . . . .
using Partial Fractions .
Exercises . . . . . . . .
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2
4
7
10
12
2
8.1
Revision of Integration
R
Recall that f (x)dx is defined as the integral of a f (x) with respect to x. An indefinite
integral has no limits. Here we must always add an arbitrary constant to the answer.
Z
f (x) dx = F (x) + C
A definite integral has limits, say we integrate x between a and b. This does not require
an arbitrary constant and can be interpreted as the area under the curve.
Z b
Area =
f (x) dx = F (b) − F (a)
a
d
F (x) = f (x). This is the Fundamental Theorem of
where F (x) is a function such that
dx
Calculus
In F11EP1 you learnt how to differentiate common functions and the rules of differentiation.
Here is a reminder.
• Integration: Common Functions and Rules
General
R
f (x)
x
n
sin ax
cos ax
eax
1
ax + b
1
2
a + x2
1
√
a2 − x2
λg(x)dx
(g(x) + h(x))
d
(f (x))
dx
xn+1
n+1
f (x)
+ C for n 6= −1
cos ax
+C
a
sin ax
+C
a
eax
+C
a
−
1
ln(ax + b) + C
a
³x´
1
tan−1
+C
a
³ xa´
sin−1
+C
a
R
λ g(x)dx
R
R
g(x)dx + h(x)dx
f (x) + C
f (x)
x
Example R
x8
8
7
sin 5x
cos(2x)
e−4x
1
3x−2
1
16 + x2
1
√
9 − x2
10e5x
x2 + cos x
d
(sin 3x)
dx
f (x)
+C
cos 5x
+C
5
sin(2x)
+C
2
e−4x
+C
−
4
−
1
ln(3x − 2) + C
3
³x´
1
tan−1
+C
4
³ x4´
sin−1
+C
3
2e5x + C
x3
+ sin x + C
3
sin 3x + C
3
• Integration: Some Practice
Z
(4e−2x + 9x2 ) dx
Example 8.1
Solution
Z
Z
Z
x3
e−2x
−2x
2
−2x
+ 9 + C = −2e−2x + 3x3 + C
+ 9x ) dx = 4 e
dx + 9 x2 dx = 4
(4e
−2
3
Z
Example 8.2
(cos(E/8) + sin(E/2)) dE
Solution
Z
sin(E/8) cos(E/2)
−
+ C = 8 sin(E/8) − 2 cos(E/2) + C
(cos(E/8) + sin(E/2)) dE =
1/8
1/2
Z
2
Example 8.3
0
Solution
Z
(6x2 − 2x) dx
2
0
Z
5
Example 8.4
1
¤2
£
(6x2 − 2x) dx = 2x3 − x2 0 = (16 − 4) − (0 − 0) = 12
3
dV
V
Solution
Z
5
1
Z
∞
Example 8.5
3
dV = [3 ln V ]51 = 3 ln(5) − 3 ln(1) = 3 ln(5)
V
100e−x dx
0
Z
Solution
∞
0
¤∞
£
100e−x dx = −100e−x 0 = (0) − (−100) = 100
Exercises 8.1
Z
Z
Z
3
3
α
1. Evaluate 2x − 16x dx 2. Evaluate (e + 3) dα 3. Evaluate −x 2 dx
Z
4. Evaluate
1
Z
7. Evaluate
1
2
3
4x − 2x dx 5. Evaluate
∞
1
dx
x2
Z
Z
2π
4
sin θ dθ 6. Evaluate
π
1
1
dx
2x
4
8.2
Integration by Substitution
Some integrals cannot be determined by just using the standard integrals above. Instead we
have to combine the standard integrals and rules with some ‘tricks’. One trick is Integration
by Substitution (which is really the opposite of the chain rule). It is easiest the understand
the method by considering an example.
Example 8.6 Find
R
e2x+1 dx
Solution Can’t do this with the standard integrals and rules learnt so far! Instead we
can use the following trick :
Trick: Introduce a new variable u, which is related to x by
u = 2x + 1
Differentiating with respect to x gives
du
=2
dx
Re-arranging (regarding du and dx as symbols) gives
du = 2 dx or dx =
1
du
2
We then substitute the expressions for u and du into the original expression
Z
Z
Z
1
2x+1
u1
eu du
e
dx =
e du =
2
2
This is now a standard integral so we can find the integral
Z
1
1
eu du = eu + C
2
2
Finally we substitute again so our answer is in terms of x (as that’s what we started with)
1 u
1
e + C = e2x+1 + C
2
2
• Summary: To apply the method of integration by substitution
1. Let u be some function of the original variable x.
2. Differentiate to find
du
.
dx
3. Rearrange to get dx in terms of du.
4. Substitute into the original integral.
5. Do integral.
6. Replace u in the answer.
5
Example 8.7 Find
Z
√
x + 1 dx
du
= 1 so du = dx. Therefore
dx
Z
Z
Z
√
√
2
u3/2
1/2
+ C = (x + 1)3/2 + C
x + 1 dx =
u du =
u du =
3/2
3
Solution
Let u = x + 1. Then
Example 8.8 Find
Z
2x
dx
+1
x2
Solution
du
du
= 2x so dx =
. Therefore
dx
2x
Z
Z
Z
2x
2x du
1
dx =
=
du
2
x +1
u 2x
u
Let u = x2 + 1. Then
Notice that the 2x term cancels on the top and the bottom. This is no accident. I chose
u = x2 +1 because I noticed its derivative was 2x which was the other term in the expression.
Continuing the solution we get
Z
1
du = ln(u) + C = ln(x2 + 1) + C
u
Example 8.9 Find
Z
6x2 cos(x3 ) dx.
Solution Let u = x3 . I have chosen this because I notice that the derivative of x3 is 3x2
du
du
= 3x2 so dx = 2 . Therefore
which is a multiple of 6x2 . Then
dx
3x
Z
Z
Z
du
6x2 cos(x2 ) dx =
6x2 cos(u) 2 =
2 cos(u) du = 2 sin(u) + C = 2 sin(x3 ) + C.
3x
• How can you determine which substitution to make?
• Experience...
...which is gained from doing exercises!
• Look for an expression and it’s derivative.
• Just try something!
6
Example 8.10 Calculate
Z
e
1
Solution
R
ln(t)/t;
ln(t)
dt
t
To calculate the definite integral we first need to know the indefinite integral
Z
ln(t)
dt
t
Guessing or noticing that 1/t is the derivative of ln(t) we put u = ln(t). Then
1
du
=
dt
t
so dt = t du
Therefore
Z
ln(t)
dt =
t
Z
u
t du =
t
Z
u du =
u2
1
+ C = (ln(t))2 + C
2
2
Having found the indefinite integral we can now compute the definite integral using the
fundamental theorem of calculus:
Z e
h1
ie
ln(t)
1
1
1
dt =
(ln(t))2 = (ln(e))2 − (ln(1))2 =
t
2
2
2
2
1
1
(n.b. ln(e) = 1 and ln(1) = 0).
Exercises 8.2
1.Z
a.
Z
d.
Z
f.
Use the given substitution toZ find each of the following indefinite
Z integrals.
3
b. sin(2θ −4) dθ; u = 2θ −4
c. x2 ex dx;
(x+7)9 dx; u = x+7
Z
x
3
sin (5t) cos(5t) dt; u = sin(5t)
e. √
dx; u = x2 + 1
2
x +1
Z
ln(x)
1−5x
dx; u = ln(x)
dx; u = 1 − 5x
g.
e
x
u = x3
2.Z Use a substitution Zto find each of the following
Z indefinite integrals.
Z
2
5
2
4x−5
a. (−x − 3) dx
b. θ sin(θ + 4) dθ
c. e
dx
d. xe2−x dx
Z
Z
Z
Z
cos(x)
x
2x + 1
3
dx
h.
dx
f. √
dx
g.
e. cos (t) sin(t) dt
2
x +x+1
sin4 (x)
1 − x2
3. Evaluate the following definite integrals.
Z 1
Z 2
Z π/2
2 x3
7
a.
x e dx
b.
(2x + 4) dx
c.
x cos(x2 ) dx
0
1
0
Z
d.
e
e2
1
dx
x ln(x)
7
8.3
Integration by Parts
Another method for integration when standard rules cannot be used is Integration by parts.
If u(x) and v(x) are two functions then
Z
Z
0
u0 (x) v(x) dx
u(x) v (x) dx = u(x) v(x) −
The above fact can be obtained from the product rule for derivatives and the definition of
indefinite integrals.
Example 8.11 Derivation of integration by parts formula.
Solution To see how the integration by parts formula can be derived, we start by recalling
the product rule (for derivatives):
d
(u(x) v(x)) = u(x) v 0 (x) + u0 (x) v(x)
dx
Integrating both sides then gives
Z
Z
d
(u(x) v(x)) dx =
(u(x) v 0 (x) + u0 (x) v(x)) dx
dx
Z
Therefore
u(x) v(x) + C =
0
Z
u(x) v (x) dx +
u0 (x) v(x) dx
Re-arranging (and absorbing the integration constant back inside the indefinite integral)
then gives the fact stated above:
Z
Z
0
u(x) v (x) dx = u(x) v(x) − u0 (x) v(x) dx
Integration by parts is used to simplify an integral; the hope is that the integral on the
RHS is simpler than the one on the LHS!
Z
Example 8.12 Find
Solution
so
ln(x) x dx
Let
u(x) = ln(x) and v 0 (x) = x
1
u (x) =
x
0
Z
and v(x) =
0
v (x) dx =
Z
x dx =
x2
(+C)
2
8
Thus
Z
x2
−
ln(x) x dx = ln(x)
2
Z
1 x2
x2 1
dx = ln(x)
−
x 2
2
2
Z
x dx = ln(x)
x2 x2
−
+C
2
4
• Summary: To apply the method of integration by parts:
1. Identify u(x) and v 0 (x).
Z
0
2. Determine u (x) and v(x) =
Z
3. Plug these into the formula
v 0 (x) dx.
0
uv = uv −
Z
u0 v
4. Do remaining integral.
Z
Example 8.13 Find
Solution
x sin(x) dx
Let
u = x and v 0 = sin(x)
u0 = 1 and v = − cos(x)
Therefore
R
x sin(x) dx
R
= x(− cos(x)) − 1(− cos(x)) dx
R
= −x cos(x) + cos(x) dx = −x cos(x) + sin(x) + C
• How do you choose u(x) and v 0 (x) ?
• Experience...
...which is gained from doing exercises!
• Remember that the integral on the right hand side should be simpler than the original
integral!
• Just try something!
In the previous example the choice
u = sin(x) and v 0 = x
is bad since
u0 = cos(x) and v =
Z
x2
2
Z
x2
x2
x sin(x) dx = sin(x)
− cos(x)
dx
2
2
The integral on the RHS is more complicated than the original integral!
so
9
Z
Example 8.14 Find
Solution
xex dx
Let
u(x) = x and v 0 (x) = ex
so
u0 (x) = 1 and v(x) = ex .
Z
Therefore
x
x
Z
xe dx = xe −
ex dx = xex − ex + C.
• Integrating by parts more than once
Sometimes it may be necessary to apply integration by parts twice (or even more!). For
example
Z
Example 8.15 Find
Solution
x2 cos(x) dx
We let
u = x2
so
and v 0 = cos(x)
u0 = 2x and v = sin(x)
Therefore
Z
2
2
x cos(x) dx = x sin(x) −
Z
2
2x sin(x) dx = x sin(x) − 2
Z
x sin(x) dx
The integral on the RHS can be done by applying integration by parts again! We did this
in an earlier example so we now get
Z
x2 cos(x) dx = x2 sin(x)−2(−x cos(x)+sin(x))+C = x2 sin(x)+2x cos(x)−2 sin(x)+C
Exercises 8.3
1.Z Find the following using integration by parts with
Z the given u and v.
b. te−4t dt; u = t, v 0 = e−4t
a. (4 − 2x) sin(x) dx; u = 4 − 2x, v 0 = sin(x)
2.Z Find the followingZusing integration byZ parts
c. xe3x−1 dx
a. x sin(2x) dx
b. x3 ln(x) dx
3.Z Use your results from
the following definate integrals.
Z 3question 2 to determine
Z 2
π
a.
x sin(2x) dx
b.
x3 ln(x) dx
c.
xe3x−1 dx
0
2
1
Z
4.
Apply integration by parts twice to evaluate
0
2
x2 ex dx
10
8.4
Integration using Partial Fractions
• Partial Fractions
Recall partial fractions are a way of splitting complicated rational expressions into simpler
parts. For example, we have
2
5
7x + 3
=
+
(x − 1)(x + 4)
x−1 x+4
To achieve this we begin by writing
A
B
7x + 3
=
+
(x − 1)(x + 4)
x−1 x+4
Multiplying both sides by (x − 1)(x + 4) gives
7x + 3 = A(x + 4) + B(x − 1)
Putting x = 1 leads to
7 × 1 + 3 = A × (1 + 4) + B × (1 − 1)
=⇒
10 = 5A
=⇒
A = 2,
while putting x = −4 gives
7 × (−4) + 3 = A × (−4 + 4) + B × (−4 − 1)
Therefore
=⇒
−25 = −5B
=⇒
B = 5.
2
5
7x + 3
=
+
.
(x − 1)(x + 4)
x−1 x+4
• Integrating using partial fractions
Partial fractions can be used to help integrate rational functions. For example, to find
Z
7x + 3
dx
(x − 1)(x + 4)
we can use the partial fraction decomposition found above to get
Z
Z
Z
7x + 3
2
5
dx =
dx +
dx
(x − 1)(x + 4)
x−1
x+4
Now
Z
2
dx =
x−1
Z
2
du
u
while
Z
Z
5
5
dx =
du
x+4
u
Hence
Z
Z
(with u = x − 1) = 2
Z
(with u = x + 4) = 2
1
du = 2 ln(u) + C = 2 ln(x − 1) + C,
u
1
du = 5 ln(u) + C = 5 ln(x + 4) + C.
u
7x + 3
dx = 2 ln(x − 1) + 5 ln(x + 4) + C
(x − 1)(x + 4)
11
• Summary: To integrate a rational function using partial fractions:
1. Write the rational function in terms of partial fractions.
2. Integrate.
Note: The second step will generally involve the integral
Z
1
du = ln(u) + C
u
Z
Example 8.16
Solution
3
dx
x(x + 3)
Firstly we find a partial fractions decomposition; suppose
A
B
3
= +
x(x + 3)
x x+3
so (multiplying by x(x + 3))
3 = A(x + 3) + Bx
Putting x = 0 leads to
3 = 3A + 0
=⇒
A=1
while putting x = −3 gives
3 = 0 + (−3)B
Therefore
so
Z
3
dx =
x(x + 3)
=⇒
B = −1
1
1
3
= −
x(x + 3)
x x+3
Z
1
dx −
x
Z
³ x ´
1
dx = ln(x) − ln(x + 3) + C = ln
+C
x+3
x+3
Exercises 8.4
1.Z Use partial fractions to find
Z the integrals below. Z
13x − 4
1
1
dx
b.
dx
c.
dx
a.
(2 + 2x)(x − 3)
6x2 − x − 2
x2 − 2x − 3
2.Z Evaluate the following integrals.
Z 4
2
1
2x + 3
a.
dx
b.
dx
1 (1 + x)(3 + x)
0 (x + 3)(2x + 5)
Z
2
c.
1
x2
1
dx
+ 2x
12
8.5
Answers to Exercises
Exercise Set 8.1
1
2 5
1. x4 − 8x2 + C 2. eα + 3α + C 3. − x 2 + C
2
5
4. 12 5. −2 6. ln(2) 7. 1
Exercise Set 8.2
3
10
1. (a)
−1/2 cos(2 θ −4)+C, (c) 1/3 ex +C, (d) 1/20 (sin(5 t))4 +C,
√ 1/10 (x + 7) +C, (b)1−5
(e) x2 + 1 + C, (f) −1/5 e x + C, (g) 1/2 (ln(x))2 + C.
2
2. (a) −1/6 (−x − 3)6 + C, (b)
−1/2 cos(θ2 + 4) +C, (c) 1/4 e4 x−5 + C, (d) −1/2 e2−x + C,
√
(e) −1/4 (cos(t))4 + C, (f) − 1 − x2 + C, (g) ln(x2 + x + 1) + C, (h) −1/3 (sin(x))−3 + C.
3. (a) 0.573 to 3 d.p., (b) 943600, (c) 0.312 to 3 d.p., (d) 0.693 to 3 d.p.
Exercise Set 8.3
1. (a) −4 cos(x) − 2 sin(x) + 2 x cos(x) + C, (b) −1/4 te−4 t − 1/16 e−4 t + C.
2. (a) 1/4 sin(2 x)−1/2 x cos(2 x)+C, (b) 1/4 x4 ln(x)−1/16 x4 +C, (c) 1/9 (3 x − 1) e3 x−1 +
C.
3. (a) −π/2, (b) 15.412 to 3 d.p., (c) 80.808 to 3 d.p.
4. 2e2 − 2 = 12.778 to 3 d.p.
Exercise Set 8.4
1. (a) −1/8 ln(2 + 2x) + 1/8 ln(x − 3) + C, (b) 3/2 ln(2 x + 1) + 2/3 ln(3 x − 2) + C,
(c) −1/4 ln(1 + x) + 1/4 ln(x − 3) + C.
2. (a) 0.0912 to 4 d.p., (b) 0.6309 to 4 d.p., (c) 0.2027 to 4 d.p.