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
With the above substitution, the given integral is given by

In what follows C is a constant of integration which is added in the final result.
Examples
Example 1
Evaluate the integral
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
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
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
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
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
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
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
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
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
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
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. ∫ e 4x - 7 dx
3. ∫ x(4x 2 + 5) 4 dx
4. ∫ 1 / (x + 3) 3 dx
Answers to Above Exercises
1. (1/3) sin(3x - 2)
2. (1/4) e 4x - 7 + C
3. (1/40) (4x 2 + 5) 5 + C
4. (-1/2) 1 / (x + 3) 2 + C