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

With the above substitution, the given integral is given by

In what follows C is a constant of integration which is added in the final result.

__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

?e^{3x - 2} dx

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

= (1/3) e^{u}

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

__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} + 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) ?(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

__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 = 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} + 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) sin^{4}(x) dx
= ?u^{4} du

= (1/5) u^{5}

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

= ?(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

__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} + 4u^{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} + C

__Solution to Example 11:__

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| + C

1. ?cos(3x - 2) dx

2. ?e

3. ?x(4x

4. ?1 / (x + 3)

1. (1/3) sin(3x - 2)

2. (1/4) e

3. (1/40) (4x

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