Question

How do you integrate `int (3-4x)^8` using substitution?

Answer 1

`int (3-4x)^(8) dx = -1/36(3-4x)^(9) + C`

Explanation:

In this problem, we can do a u-substitution.

Let `u = 3-4x`, then `du = -4 dx -> -1/4 du = dx`

Rewriting our integral and substituting back our `u` we have

`int (3-4x)^(8) dx = -1/4 int u^8 du =-1/4 int u^8 du`

`= -1/4 * 1/9u^(9) + C`

`= -1/36(3-4x)^(9) + C`