INTEGRATION BY SUBSTITUTION
• "Millions saw the apple fall, but
Newton asked why." -– Bernard Baruch
integration by using u-substitution
&
integration by using trig substitution
Integration by using u-substitution
Guidelines for making a change of variables
1. Choose a 𝑢 = 𝑢(𝑥)
2. Compute 𝑑𝑢 =
𝑑𝑢
𝑑𝑥
𝑑𝑥
3. Rewrite the integral in terms of 𝑢
4. Evaluate the integral in terms of 𝑢
5. Replace 𝑢 by 𝑢(𝑥)
6. Check your answer by differentiating
 x
2
 2 x  dx
1
2
u  x2 1
du  2 xdx
u3
  u  du   C
3
2

x

2

3
1
C
3
Evaluate
 5 cos 5 x dx
u  5x
du  5dx
du
  5 cos u
5
 sin u  C  sin 5x  C
Evaluate
 x x
 1 dx
2
u  x2 1
du  2xdx
2
1
  u  du
2
3
2
u3
x  1  C
 C 
6
6

Evaluate
2

u  2x 1
du  2dx
2 x 1dx
 u
du
du
  u1 2
2
2
1 2u 3 2
2x 1

C 
2 3
3
32
C
Substitution and the General Power Rule
What would you let u = in the following examples?
4
3
(
3
x

1
)
dx

2



2
x

1
x
 x  dx

2
3
3
x
x
 2 dx

 4x
dx
2
1 2x


2
cos
 x sin x dx
𝑢 = 3𝑥 − 1
𝑢 = 𝑥2 + 𝑥
𝑢 = 𝑥3 − 2
𝑢 = 1 – 2𝑥2
𝑢 = cos 𝑥
Evaluate
 (2 x
3
 3) (6 x )dx
20
2
u  (2 x 3  3)dx
du  6 x 2 dx
u 21
(2 x 3  3) 21
  u du 
C 
C
21
21
20
Evaluate
x
 x 2  9 dx  ?
1
 ln x 2  9  C
2
u  x2  9
Evaluate
1
 4  x 2 2 xdx  ?
𝑢 = 4 + 𝑥2
 ln 4  x 2  C
Evaluate
2
4
(
x

3
x

7
)
(2 x  3)dx  ?

𝑢 = 𝑥 2 − 3𝑥 + 7
( x 2  3 x  7)5

C
5
Evaluate
3
4
2
(
2
x

3
)
(
6
x
)dx  ?

(2 x 3  3)5

C
5
u  2 x3  3
Evaluate
2
3
x
sin
x
dx

𝑢 = 𝑥3
𝑑𝑢 = 3𝑥2 𝑑𝑥
1
  sin u du
3
du
1
1
 dx
3
2
 ( cos u )  C   cos x  C 3x
3
3
Evaluate
2
sin
 3x cos 3x dx
3
1 2
u3
sin
3x
  u du   C 
C
3
9
9
 u  1  u du

Evaluate  x 2 x 1dx
  2  2


1
  u 3 2  u1 2 du
4
𝑢 = sin 3𝑥
𝑑𝑢 = 3cos 3𝑥 𝑑𝑥
𝑢 = 2𝑥 − 1
𝑑𝑢 = 2𝑑𝑥
52
32


2 x  1
2 x  1
1  2u 5 2 2u 3 2 
  C 

C
 

10
6
4 5
3 
Evaluate
 tan x dx
sin x
 tan x dx   cos x dx u  cos x
1
   du   ln u  C
u
  ln cos x  C
 ln (cos x)
1
1
 C  ln
cos x
 ln sec x  C
C
1
Evaluate
 x( x
2
 1) dx
3
𝑢 = 𝑥2 + 1
𝑑𝑢 = 2𝑥 𝑑𝑥
0
1 3
  u du
2
Note that there are no upper and lower
limits of integration.
u4  2
 
81
We must determine new upper and
lower limits by substituting the old
ones in for 𝑥 in 𝑢 = 𝑥2 + 1
16 1 15
  
8 8 8
Or, we could use the old limits if we
substitute 𝑥2 + 1 back in.


1

x 1


8  0
2
4
16 1 15
  
8 8 8
5
Evaluate
u  2x 1
x
dx
2x 1

1
u 2  2x 1
u  1 udu 
  (u 2  1) du

2u
1
1
3
3
2
3

1u
   u 
2 3
 2
3
1  27   1 
16
   3     1 
3
2  3
  3 
2𝑢 𝑑𝑢 = 2 𝑑𝑥
𝑥
u  2x 1
1
1
5
3
Integration of Even and Odd Functions
If f is an even function, then
2
a
a
a
0
 f ( x)dx  2 f ( x)dx
Ex.

0
x
dx

3
2
Odd or Even?
If f is an odd function, then
a
 f ( x)dx  0
a
Trigonometric substitution
The following examples and exercises are difficult.
I hope you are unlikely to be asked to do them in an
exam but you will find it useful to follow the method.
Certain integrals can be reduced to easier forms by an
appropriate trigonometric substitution.
We use three important identities in trigonometry in this lesson.
(i)
1 – sin2  = cos2 
(ii)
sec2  – 1= tan2 
(iii)
1 + tan2  = sec2 
Substitution
Form of
expression
𝑥 = 𝑎 𝑠𝑖𝑛𝜃
𝑎2 − 𝑥 2 =
𝑎2 − 𝑎2 𝑠𝑖𝑛2 𝜃 = 𝑎 cos 𝜃
𝑥 = 𝑎 𝑠𝑒𝑐𝜃
𝑥 2 − 𝑎2 =
𝑎2 𝑠𝑒𝑐 2 𝜃 − 𝑎2 = 𝑎 tan 𝜃
𝑥 = 𝑎 𝑡𝑎𝑛𝜃
𝑎2 + 𝑥 2 =
𝑎2 + 𝑎2 𝑡𝑎𝑛2 𝜃 = 𝑎 sec 𝜃
Evaluate

1
1 x
2
x  sin u
dx
dx  cos u du
N.B. Instead of defining u as a function of x we have defined x as a function of u

1
1 x
2
dx 


1
cos u du
1  sin u
1
cos u du   1 du
cos 2 u
2
 u  C  sin 1 x  C
OR
y  sin

1
1
1 x2
dy
x 

dx
1
1 x2
dx  sin 1 x  C
Evaluate


1
9 x
2
1
9  x2
dx
x  3sin u
dx 
1



9  9 sin 2 u
1
9(1  sin 2 u )
1
9 cos 2 u
3 cos u du
3 cos u du
3 cos u du
  1 du  u  C
 x
 sin    C
3
1
dx  3 cos u du
1
1
1 x
dx  tan
Evaluate 
2
4 x
2
2
x  2 tan u
dx  2 sec 2 u du
1
1
2
dx

2
sec
u du
 4  x2
 4  4 tan 2 u
1
2

2
sec
u du
2
4(1  tan u )
sec 2 u  1  tan 2 u
1
1
  du  u  C
2
2
1
1 x
 tan
C
2
2
Evaluate
1
𝑥2
25 −
𝑥 = 5 𝑠𝑖𝑛
𝑥2
𝑑𝑥 = 5 𝑐𝑜𝑠  𝑑
1
=
25
𝑠𝑖𝑛2 𝜃
25 − 25
5cos θ
=
25 𝑠𝑖𝑛2 𝜃 25 𝑐𝑜𝑠 2 𝜃
=
1
25
𝑐𝑠𝑐 2 𝜃 𝑑𝜃 =
5cos θ 𝑑𝜃
𝑑𝜃 =
1
(− cot 𝜃)
25
1 cos 𝜃
1
=−
=−
25 sin 𝜃
25
=−
𝑠𝑖𝑛2 𝜃
25 − 𝑥 2
+𝐶
25𝑥
1
𝑑𝜃
25 𝑠𝑖𝑛2 𝜃
+𝐶
1 − 𝑠𝑖𝑛2 𝜃
1
=−
sin 𝜃
25
𝑥
1−
5
𝑥
5
2