Integral Calculus
Assumed Knowledge
Reminder — Standard Integrals
Note — Four Basic Rules
Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x , x1 and sec2 x
Example — Standard Integrals of e x , x1 and sec2 x
Integration by Substitution
Definition — Differentials
Note — Integration by Substitution
Example — Integration by Substitution
5.1 Assumed Knowledge
Reminder — Standard Integrals
5.1 Assumed Knowledge
Reminder — Standard Integrals
I
R
ax n dx =
a
x n+1 + c.
n+1
5.1 Assumed Knowledge
Reminder — Standard Integrals
I
I
a
x n+1 + c.
n+1
R
1
cos(ax + b) dx = sin(ax + b) + c.
a
R
ax n dx =
5.1 Assumed Knowledge
Reminder — Standard Integrals
I
I
I
a
x n+1 + c.
n+1
R
1
cos(ax + b) dx = sin(ax + b) + c.
a
R
1
sin(ax + b) dx = − cos(ax + b) + c.
a
R
ax n dx =
5.1 Assumed Knowledge
Reminder — Standard Integrals
a
x n+1 + c.
n+1
R
1
I
cos(ax + b) dx = sin(ax + b) + c.
a
R
1
I
sin(ax + b) dx = − cos(ax + b) + c.
a
The notation F(x) is Rused to represent the antiderivative of f (x) so
that F0 (x) = f (x) or f (x) dx = F(x) + c.
I
R
ax n dx =
Note — Four Basic Rules
Note — Four Basic Rules
1.
R
R
R
R
(af (x)) + (bg (x)) dx = a f (x) dx + b g (x) dx.
Note — Four Basic Rules
1.
R
R
R
R
(af (x)) + (bg (x)) dx = a f (x) dx + b g (x) dx.
2.
Rb
a
f (x) dx = F(b) − F(a) where F0 (x) = f (x).
Note — Four Basic Rules
1.
R
R
R
R
(af (x)) + (bg (x)) dx = a f (x) dx + b g (x) dx.
2.
Rb
3.
Rc
a
a
f (x) dx = F(b) − F(a) where F0 (x) = f (x).
f (x) dx =
Rb
a
f (x) dx +
Rc
b
f (x) dx, where a < b < c.
Note — Four Basic Rules
1.
R
R
R
R
(af (x)) + (bg (x)) dx = a f (x) dx + b g (x) dx.
2.
Rb
3.
Rc
4.
Rb
a
a
a
f (x) dx = F(b) − F(a) where F0 (x) = f (x).
f (x) dx =
Rb
a
f (x) dx = −
f (x) dx +
Ra
b
f (x) dx.
Rc
b
f (x) dx, where a < b < c.
Note — Four Basic Rules
1.
R
R
R
R
(af (x)) + (bg (x)) dx = a f (x) dx + b g (x) dx.
2.
Rb
3.
Rc
4.
Rb
a
a
a
f (x) dx = F(b) − F(a) where F0 (x) = f (x).
f (x) dx =
Rb
a
f (x) dx = −
f (x) dx +
Ra
b
Rc
b
f (x) dx, where a < b < c.
f (x) dx.
Item 2 is often referred to as the Fundamental Theorem of
Calculus. The integral represents the area under the curve f (x)
between the points x = a and x = b.
5.2 Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x ,
1
x
and sec2 x
5.2 Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x ,
I
R
e x dx = e x + c.
1
x
and sec2 x
5.2 Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x ,
I
R
e x dx = e x + c.
I
R
1
x
dx = ln |x| + c.
1
x
and sec2 x
5.2 Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x ,
I
R
e x dx = e x + c.
I
R
1
x
I
R
sec2 x dx = tan x + c.
dx = ln |x| + c.
1
x
and sec2 x
5.2 Standard Integrals of e x ,
1
x
and sec2 x
Definition — Standard Integrals of e x ,
I
R
e x dx = e x + c.
I
R
1
x
I
R
sec2 x dx = tan x + c.
1
x
and sec2 x
dx = ln |x| + c.
Note |x| is the modulus of x and |x| = x, when x ≥ 0 but
|x| = −x, when x ≤ 0.
Example — Standard Integrals of e x ,
1
x
and sec2 x
Example — Standard Integrals of e x ,
1
x
and sec2 x
Find the following indefinite and definite integrals.
R
1. (e 2x + 5x) dx
Example — Standard Integrals of e x ,
1
x
and sec2 x
Find the following indefinite and definite integrals.
R
1. (e 2x + 5x) dx
R
R
= e 2x dx + 5x dx
Example — Standard Integrals of e x ,
1
x
and sec2 x
Find the following indefinite and definite integrals.
R
1. (e 2x + 5x) dx
R
R
= e 2x dx + 5x dx
= 12 e 2x + 25 x 2 + c
Example — Standard Integrals of e x ,
1
x
and sec2 x
Find the following indefinite and definite integrals.
R
1. (e 2x + 5x) dx
R
R
= e 2x dx + 5x dx
= 12 e 2x + 25 x 2 + c
2.
R
1
dx
5x + 4
Example — Standard Integrals of e x ,
1
x
and sec2 x
Find the following indefinite and definite integrals.
R
1. (e 2x + 5x) dx
R
R
= e 2x dx + 5x dx
= 12 e 2x + 25 x 2 + c
2.
R
1
dx
5x + 4
=
1
5
ln |5x + 4| + c
3.
R
π
4
0
1 + sin2 x
cos2 x
dx
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
0
sec2 x dx +
R
π
4
0
tan2 x dx
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
=
R
π
4
0
0
sec2 x dx +
R
π
4
sec2 x dx +
R
π
4
0
0
tan2 x dx
sec2 x − 1 dx
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
=
R
π
4
0
0
=2
R
sec2 x dx +
R
π
4
sec2 x dx +
R
π
4
sec2 x dx −
R
π
4
0
0
0
tan2 x dx
sec2 x − 1 dx
π
4
0
1 dx
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
=
R
π
4
0
0
=2
R
sec2 x dx +
R
π
4
sec2 x dx +
R
π
4
sec2 x dx −
R
π
4
0
π
0
π
= 2[tan x]04 − [x]04
0
tan2 x dx
sec2 x − 1 dx
π
4
0
1 dx
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
=
R
π
4
0
0
=2
R
sec2 x dx +
R
π
4
sec2 x dx +
R
π
4
sec2 x dx −
R
π
4
0
π
0
0
tan2 x dx
sec2 x − 1 dx
π
4
0
1 dx
π
= 2[tan x]04 − [x]04
= 2(tan π4 − tan 0) − ( π4 − 0)
1 + sin2 x
dx
3. 0
cos2 x
Rπ
R π4 sin2 x
1
= 04
dx
+
dx
0
cos2 x
cos2 x
R
π
4
=
R
π
4
=
R
π
4
0
0
=2
R
sec2 x dx +
R
π
4
sec2 x dx +
R
π
4
sec2 x dx −
R
π
4
0
π
0
0
tan2 x dx
sec2 x − 1 dx
π
4
0
1 dx
π
= 2[tan x]04 − [x]04
= 2(tan π4 − tan 0) − ( π4 − 0)
=2−
π
4
Further Examples
Maths In Action: Book 1
Page 72
Exercise 2A/2B
5.3 Integration by Substitution
Definition — Differentials
5.3 Integration by Substitution
Definition — Differentials
For y = f (x), if δy is a small change in y induced by a small
change in x, δx, then
δy = lim
δx→0
i.e.
δy =
δy
.δx
δx
dy
.δx
dx
5.3 Integration by Substitution
Definition — Differentials
For y = f (x), if δy is a small change in y induced by a small
change in x, δx, then
δy = lim
δx→0
i.e.
δy =
δy
.δx
δx
dy
.δx
dx
The smaller δx becomes, the better the approximation for δy .
5.3 Integration by Substitution
Definition — Differentials
For y = f (x), if δy is a small change in y induced by a small
change in x, δx, then
δy = lim
δx→0
i.e.
δy =
δy
.δx
δx
dy
.δx
dx
The smaller δx becomes, the better the approximation for δy .
dy is known as the y -differential and dx is known as the
dy
x-differential.
is the coefficient of dx and is referred to as the
dx
differential coefficient.
Note — Integration by Substitution
Note — Integration by Substitution
When differentiating a composite function y = g (f (x)) using the
dy
chain rule, then
= g 0 (f (x)).f 0 (x).
dx
Note — Integration by Substitution
When differentiating a composite function y = g (f (x)) using the
dy
chain rule, then
= g 0 (f (x)).f 0 (x).
dx
Let u = f (x) then using differentials du = f 0 (x) dx therefore
Z
Z
g 0 (f (x)).f 0 (x) dx = g 0 (u) du.
Note — Integration by Substitution
When differentiating a composite function y = g (f (x)) using the
dy
chain rule, then
= g 0 (f (x)).f 0 (x).
dx
Let u = f (x) then using differentials du = f 0 (x) dx therefore
Z
Z
g 0 (f (x)).f 0 (x) dx = g 0 (u) du.
R
Hopefully g 0 (u) du is a standard form and integration can be
performed.
Note — Integration by Substitution
When differentiating a composite function y = g (f (x)) using the
dy
chain rule, then
= g 0 (f (x)).f 0 (x).
dx
Let u = f (x) then using differentials du = f 0 (x) dx therefore
Z
Z
g 0 (f (x)).f 0 (x) dx = g 0 (u) du.
R
Hopefully g 0 (u) du is a standard form and integration can be
performed.
The function f (x) is known as the essential function.
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x2 + 5
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x 2 + 5 ⇒ du = 2x
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x 2 + 5 ⇒ du = 2x
R
R
3x(x 2 + 5)7 dx = 23 u 7 du
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x 2 + 5 ⇒ du = 2x
R
R
3x(x 2 + 5)7 dx = 23 u 7 du
=
3
2
× 18 u 8 + c
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x 2 + 5 ⇒ du = 2x
R
R
3x(x 2 + 5)7 dx = 23 u 7 du
× 18 u 8 + c
=
3
2
=
3 8
16 u
+c
Example — Integration by Substitution
1. Find
R
3x(x 2 + 5)7 dx
u = x 2 + 5 ⇒ du = 2x
R
R
3x(x 2 + 5)7 dx = 23 u 7 du
× 18 u 8 + c
=
3
2
=
3 8
16 u
=
3
2
16 (x
+c
+ 5)8 + c
2. Find
R
sin x cos4 x dx
2. Find
R
sin x cos4 x dx
u = cos x
2. Find
R
sin x cos4 x dx
u = cos x ⇒ du = − sin x
2. Find
R
sin x cos4 x dx
u = cos x ⇒ du = − sin x
R
R
sin x cos4 x dx = − u 4 du
2. Find
R
sin x cos4 x dx
u = cos x ⇒ du = − sin x
R
R
sin x cos4 x dx = − u 4 du
= − 15 u 5 + c
2. Find
R
sin x cos4 x dx
u = cos x ⇒ du = − sin x
R
R
sin x cos4 x dx = − u 4 du
= − 15 u 5 + c
= − 15 cos5 x + c
3. Find
R√
1 − x 2 dx
3. Find
R√
1 − x 2 dx
u = sin−1 x
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x)
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
R
= 12 cos 2u − 12 du
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
R
= 12 cos 2u − 12 du
=
1
4
sin 2u − 12 u + c
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
R
= 12 cos 2u − 12 du
=
1
4
sin 2u − 12 u + c
=
1
4
× 2 sin u cos u − 12 u + c
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
R
= 12 cos 2u − 12 du
=
1
4
× 2 sin u cos u − 12 u + c
p
= 12 sin u (1 − sin u) − 12 u + c
=
1
4
sin 2u − 12 u + c
3. Find
R√
1 − x 2 dx
u = sin−1 x ⇒ x = sin(x) ⇒ dx = cos u du
R√
Rp
1 − x 2 dx =
1 − sin2 u cos u du
R
= cos2 u du
R
= 12 cos 2u − 12 du
=
1
4
× 2 sin u cos u − 12 u + c
p
= 12 sin u (1 − sin u) − 12 u + c
p
= 12 x (1 − x 2 ) − 12 sin−1 x + c
=
1
4
sin 2u − 12 u + c
4. Find
R
1√
1− x
dx
(x 6= 1)
4. Find
R
1√
1− x
u =1−
√
x
dx
(x 6= 1)
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2 ⇒ dx = −2(1 − u)du
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2 ⇒ dx = −2(1 − u)du
R 1
R −2(1−u)
√ dx =
du
u
1− x
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2 ⇒ dx = −2(1 − u)du
R 1
R −2(1−u)
√ dx =
du
u
1− x
R
= (−2u −1 + 2) du
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2 ⇒ dx = −2(1 − u)du
R 1
R −2(1−u)
√ dx =
du
u
1− x
R
= (−2u −1 + 2) du
= −2 ln |u| + 2u + c
4. Find
R
1√
1− x
u =1−
√
dx
(x 6= 1)
1
x ⇒ du = − 21 x − 2 dx
1
Therefore dx = −2x 2 ⇒ dx = −2(1 − u)du
R 1
R −2(1−u)
√ dx =
du
u
1− x
R
= (−2u −1 + 2) du
= −2 ln |u| + 2u + c
√
√
= −2 ln |1 − x| + 2(1 − x) + c
Further Examples
Maths In Action: Book 1
Page 74
Exercise 3/4A/4B