Integration by Substitution

In this section we reverse the Chain rule of differentiation and derive a method for solving integrals
called the method of substitution. Recall the chain rule of differentiation says that
d
f (g(x)) = f 0 (g(x))g 0 (x).
dx
Reversing this rule tells us that
Z
f 0 (g(x))g 0 (x) dx = f (g(x)) + C

Example Use the chain rule to find the derivative of the composite function f (g(x)) = (x2 + 1)2 and
identify f and g in the expression.

f 0 (g(x))g 0 (x) dx and evaluate it :

R
Write the integral below as
Z
4x(x2 + 1) dx

The Substitution Rule says that if g(x) is a differentiable function whose range is the interval I
and f is continuous on I, then
Z Z
0
f (g(x))g (x) dx = f (u) du

where u = g(x) and du = g 0 (x) dx.

When applying the method, we substitute u = g(x), integrate with respect to the variable u and then
reverse the substitution in the resulting antiderivative.
R √
Example Find 2x x2 + 1 dx.
R √ Rp
Here we let g(x) = x2 + 1. We have 2x x2 + 1 dx = g(x)g 0 (x) dx.
2 du
Now we let u = g(x) = x + 1, giving us that dx = 2x, giving us that du = 2xdx. Therefore, we have
Z √ Z
√ 2u3/2
2
2x x + 1 dx = u du = + C.
3
We convert our answer back to an answer in terms of the variable x, to get
Z √ 2u3/2 2(x2 + 1)3/2
2x x2 + 1 dx = +C = + C.
3 3
√
You should check that this the general antiderivative for 2x x2 + 1, by differentiating it using the chain
rule.

1
 R
Sometimes your substitution may result in an integral of the form f (u)c du for some constant c, which
is not a problem.
Example Find the following:
Z √ Z Z
x3 x4 + 1 dx, 3
sin x cos x dx, x sin(x2 + 3) dx

Sometimes the appropriate substitution

R is non-obvious and you may have to work a little harder to put
the resulting integral in the form f (u)du:
Example Find the following :
x3
Z
√ dx
x2 + 1

2
 The Definite Integral
The Substitution Rule For Definite Integrals If g 0 is continuous on [a, b] and f is continuous
on the range of u = g(x), then
Z b Z g(b)
0
f (g(x))g (x) dx = f (u) du.
a g(a)

Proof If F is an antiderivative for f , we have

Z b Z b b
0
f (g(x))g (x) dx = F 0 (g(x))g 0 (x) dx = F (g(x)) = F (g(b)) − F (g(a)).

a a a

On the other hand, letting u = g(x), we have

Z g(b) g(b)
f (u)du = F (u) = F (g(b)) − F (g(a))

g(a) g(a)

This gives us two options for calculating a definite integral using substitution:

1. We can calculate the antiderivative in terms of x and use the original limits of integration to
evaluate the definite integral or

2. we can change the limits of integration when we make the substitution, calculate the antiderivative
in terms of u and evaluate using the new limits of integration.

Example Evaluate the following definite integral using both methods

Z 1 √
2x x2 + 1 dx
0

Z √ 2(x2 + 1)3/2
Method 1 In our example above, we calculated 2x x2 + 1 dx = + C. Using the
3
fundamental theorem of calculus, we get
√
Z 1 √ 2(x 2
+ 1) 3/2 1
2(2)3/2
− 2(1)3/2
4 2−2
2x x2 + 1 dx =  = = .

0 3 0 3 3

Method 2 As in the example above, we substitute u = x2 + 1. When we change the variable, we also
change the limits of integration. When x = 0, u = u(x) = u(0) = 1, when x = 1, u = u(x) = u(1) = 2.
Our transformed integral is now given by
Z 1 √ Z 2 √
√ 2u3/2 2
2(2)3/2
− 2(1)3/2
4 2−2
2x x2 + 1 dx = u du =  = = .

0 1 3 1 3 3

3
Example Evaluate the following definite integrals:
π2 √
Z
sin x
Z 3 √
√ dx, x x2 + 1 dx.
0 x 2

Even and Odd Functions

Sometimes we can use symmetry to make evaluation
Ra of integrals
R a easier:
If f is an even function (f (x) = f (−x)), then −a f (x)dx = 2 0 f (x)dx.
Ra
If f is an odd function (f (x) = −f (−x)), then −a f (x)dx = 0
Example Evaluate the following definite integrals:
π
Z
4
Z 1
5
tan xdx, x4 + x2 + 1dx.
−π
4
−1

4
 Extra Examples (Please attempt these before you check the solutions)
Example Find the following indefinite integrals:
Z Z
x
√ dx, sin(2x + 1) dx
x2 + 1

Example (tricky - ish) Find the following :

Z
sin2 x cos3 x dx

Example Evaluate the following definite integrals:

Z π Z 2
3
3 x
sin θ cos θ dθ, √ dx (Use results from previous example)
π 2
x +1
4
1

5
 Extra Examples Solutions
Example Find the following indefinite integrals:
Z Z
x
√ dx, sin(2x + 1) dx
x2 + 1

Ex 1. Z
x
√ dx
2
x +1
Let u = x2 + 1,
du
du = 2xdx → xdx = 2
.
√
Z
x
Z
1 du 1
Z
1 1 u √
√ dx = √ = √ du = + C = x2 + 1 + C
x2 + 1 u 2 2 u 2 (1/2)

Ex 2. Z
sin(2x + 1) dx

Let u = 2x + 1,
du
du = 2dx → dx = 2
.

− cos u
Z Z Z
du 1
sin(2x + 1) dx = sin(u) = sin u du = +C
2 2 2

Example (tricky - ish) Find the following :

Z
sin2 x cos3 x dx

We let u = sin x and replace the extra cos2 x by 1 − u2 . We get du = cos x dx and

u3 u5 sin3 x sin5 x
Z Z
sin x cos x dx = u2 (1 − u2 ) du =
2 3
− +C = − +C
3 5 3 5

6
Example Evaluate the following definite integrals:
π
Z Z 2
3
3 x
sin θ cos θ dθ, √ dx (Use results from previous example)
π
4
1 x2 + 1

Ex 1: Z π
3
sin3 θ cos θ dθ
π
4

Let u = sin θ,
then du = cos θ dθ.
Changing the limits, we get
u( π4 ) = sin π4 = √12
√
3
u( π3 ) = sin π3 = 2

π
√  √3 " √ #
u( π3 ) 3 2
u4  1 ( 3)4
Z Z Z
3 2 1
(sin θ)3 cos θ dθ = u3 du = 3
u du =  = − √
π
4
u( π4 ) √1 41 4 16 ( 2)4
2 √
2

1h 9 1i 1h 5 i 5
= − = = .
4 16 4 4 16 64

Ex. 2 (using method 1): Above, we saw that

Z
x √
√ dx = x2 + 1 + C
x2 + 1
So 2
Z 2
x √  √ √ √ √
√ dx = x2 + 1 = 4 + 1 − 1 + 1 = 5 − 2.

1 x2 + 1 
1