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  lnu)
= (3/16)(4x + 1  ln4x + 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
(x + 2)^{3}(x + 4)^{2} dx
= u^{3}(u + 2)^{2} du
= (u^{5} + 4u^{4} + 4uu^{3})du
= (1/6) u^{6} + (4/5)u^{5} + u^{4}
= (1/6)(x + 2)^{6} + (4/5)(x + 2)^{5} + (x + 2)^{4} + K
Example 11: Evaluate the integral
( (2x + 3) / (x^{2} + 3x + 1) ) dx
Solution to Example 1:
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
= lnu
= 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.
