Integration by Substitution

Tutorials with examples and detailed solutions and exercises with answers on how to use the powerful technique of integration by substitution to find integrals.

Review Integration by Substitution

The method of integration by substitution may be used to easily compute complex integrals. Let us examine an integral of the form
ab f(g(x)) g'(x) dx
Let us make the substitution u = g(x), hence du/dx = g'(x) and du = g'(x) dx
With the above substitution, the given integral is given by
ab f(g(x)) g'(x) dx = g(a)g(b) f(u) du

In what follows C is a constant of integration which is added in the final result.

Examples

Example 1

Evaluate the integral
sin(a x + b) dx

Solution to Example 1:
Let u = a x + b which gives du/dx = a or dx = (1/a) du. The substitution helps in computing the integral as follows
sin(a x + b) dx
= (1/a) sin(u) du
= (1/a) (-cos(u)) + C
= - (1/a) cos(a x + b) + C

Example 2

Evaluate the integral
e3x - 2 dx

Solution to Example 2:
Let u = 3x - 2 which gives du/dx = 3 or dx = (1/3) du. Hence
e3x - 2 dx
= eu (1/3) du
= (1/3) eu
= (1/3) e3x - 2 + C

Example 3

Evaluate the integral
x (2x2 + 5)4 dx

Solution to Example 3:
Let u = 2x2 + 5 which gives du/dx = 4x, du = 4x dx, (1/4) du = x dx . The substitution gives
x (2x2 + 5)4 dx
= (1/4) (u)4 du
= (1/4) (1/5) u5
= (1/20) (2x2 + 5)5 + C

Example 4

Evaluate the integral
x √(2x + 1) dx

Solution to Example 4:
Let u = 2x + 1 which gives du/dx = 2 and dx = (1/2) du. Solve u = 2x + 1 for x to obtain x = (1/2)(u - 1). The substitution gives
x √(2x + 1) dx
= (1/2)(u - 1)(u)1/2 (1/2) du
= (1/4) (u3/2 - u1/2) du
= (1/4)( (2/5)u5/2 - (2/3)u3/2 )
= ( (2x + 1)3/2(3x - 1) ) / 15 + C

Example 5

Evaluate the integral
(x - 5)-4 dx

Solution to Example 5:
Let u = x - 5 which gives du/dx = 1. Substituting into the given integral, we obtain
(x - 5)-4 dx
= u-4 du
= (-1/3)u-3
= (-1/3)(x - 5)-3 + C

Example 6

Evaluate the integral
-x ex2 + 2 dx

Solution to Example 6:
Let u = x2 + 2 which gives du/dx = 2x and (1/2) du = x dx. A substitution into the given the integral gives
-x ex2 + 2 dx
= - eu (1/2) du
= - (1/2) eu du
= - (1/2) eu
= - (1/2) ex2 + 2 + C

Example 7

Evaluate the integral
cos(x) sin4(x) dx

Solution to Example 7:
Let u = sin(x) which gives du/dx = cos(x) or cos(x) dx = du. Substitute into the integral to obtain
cos(x) sin4(x) dx = u4 du
= (1/5) u5
= (1/5) sin5(x) + C

Example 8

Evaluate the integral
(3x / (4x + 1)) dx

Solution to Example 8:
Let u = 4x + 1 which gives du/dx = 4 or dx = (1/4) du. Solve u = 4x + 1 for x to obtain x = (1/4)(u - 1). Substitute to obtain
(3x / (4x + 1)) dx
= 3 (1/4) ( (u - 1) / u ) (1/4) du
= (3/16)(u - 1) / u du
= (3/16)(1 - 1/u) du
= (3/16)(u - ln|u|)
= (3/16)(4x + 1 - ln|4x + 1|) + C

Example 9

Evaluate the integral
(x / √(x - 2)) dx

Solution to Example 9:
Let u = x - 2 which gives du/dx = 1, dx = du and x = u + 2. Substitution
(x / √(x - 2)) dx
= (u+2) / √u) dx
= (u1/2 + 2u-1/2) dx
= (2/3)u3/2 + 2 * 2 u1/2
= (2/3)(x - 2)3/2 + 4(x - 2)1/2 + C

Example 10

Evaluate the integral
(x + 2)3(x + 4)2 dx

Solution to Example 10:
Let u = x + 2 which gives du/dx = 1, dx = du and also x = u - 2. Using the above substitution we obtain
(x + 2)3(x + 4)2 dx
= u3(u + 2)2 du
= (u5 + 4u4 + 4u3)du
= (1/6) u6 + (4/5)u5 + u4
= (1/6)(x + 2)6 + (4/5)(x + 2)5 + (x + 2)4 + C

Example 11

Evaluate the integral
( (2x + 3) / (x2 + 3x + 1) ) dx

Solution to Example 11:
Let u = x2 + 3x + 1 which gives du/dx = 2x + 3 or (2x + 3) dx = du. The substitution helps in computing the integral as follows
( 1 / u ) du
= ln|u|
= ln |x2 + 3x + 1| + C

Exercises

Use the table of integrals and the method of integration by parts to find the integrals below. [Note that you may need to use the method of integration by parts more than once].
1. cos(3x - 2) dx
2. e4x - 7 dx
3. x(4x2 + 5)4 dx
4. 1 / (x + 3)3 dx

Answers to Above Exercises


1. (1/3) sin(3x - 2)
2. (1/4) e4x - 7 + C
3. (1/40) (4x2 + 5)5 + C
4. (-1/2) 1 / (x + 3)2 + C

More References and links

integrals and their applications in calculus.