Tutorials with examples and detailed solutions and exercises with answers on how to use the powerful technique of integration by substitution to find integrals.
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
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
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
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
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
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
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
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
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
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
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