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

Examples

Example 1

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

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} + C

Example 3

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

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 + C

Example 5

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

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 e^{x2 + 2} dx
= ∫- e^u (1/2) du
= - (1/2) ∫e^u du
= - (1/2) e^u
= - (1/2) e^{x2 + 2} + C

Example 7

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

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

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 + C

Example 10

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

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

Answers to Above Exercises

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