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:

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 K is a constant of integration which is added in the final result.

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)

= (1/a) cos(a x + b) + K



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 + K


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 + K


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 + K


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 intergral, we obtain

(x - 5)-4 dx

= u-4 du

= (-1/3)u-3

= (-1/3)(x - 5)-3 + K


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 + K


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) + k


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|) + K


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 + k


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 + 4uu3)du

= (1/6) u6 + (4/5)u5 + u4

= (1/6)(x + 2)6 + (4/5)(x + 2)5 + (x + 2)4 + K


Example 11: Evaluate the integral

( (2x + 3) / (x2 + 3x + 1) ) dx

Solution to Example 1:

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| + K

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 + K

3. (1/40) (4x2 + 5)5 + K

4. (-1/2) 1 / (x + 3)2 + K

More references on integrals and their applications in calculus.