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

_{a}^{b} 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

_{a}^{b} 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

e^{3x - 2} dx

Solution to Example 2:

Let u = 3x - 2 which gives du/dx = 3 or dx = (1/3) du. Hence

e^{3x - 2} dx

= e^{u} (1/3) du

= (1/3) e^{u}

= (1/3) e^{3x - 2} + K

Example 3: Evaluate the integral

x (2x^{2} + 5)^{4} dx

Solution to Example 3:

Let u = 2x^{2} + 5 which gives du/dx = 4x, du = 4x dx, (1/4) du = x dx . The substitution gives

x (2x^{2} + 5)^{4} dx

= (1/4) (u)^{4} du

= (1/4) (1/5) u^{5}

= (1/20) (2x^{2} + 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) (u^{3/2} - u^{1/2}) du

= (1/4)( (2/5)u^{5/2} - (2/3)u^{3/2} )

= ( (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 e^{x2 + 2} dx

Solution to Example 6:

Let u = x^{2} + 2 which gives du/dx = 2x and (1/2) du = x dx. A substitution into the given the integral gives

-x e^{x2 + 2} dx

= - e^{u} (1/2) du

= - (1/2) e^{u} du

= - (1/2) e^{u}

= - (1/2) e^{x2 + 2} + K

Example 7: Evaluate the integral

cos(x) sin^{4}(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) sin^{4}(x) dx
= u^{4} du

= (1/5) u^{5}

= (1/5) sin^{5}(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

= (u^{1/2} + 2u^{-1/2}) dx

= (2/3)u^{3/2} + 2 * 2 u^{1/2}

= (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

Let u = x^{2} + 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 |x^{2} + 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. 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} + K

3. (1/40) (4x^{2} + 5)^{5} + K

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

More references on
integrals and their applications in calculus.