Question

How do you integrate `int sinx/(cosx)^5` using substitution?

Answer 1

The answer is `=1/(4cos^4x)+C`

Explanation:

Let `u=cosx`

`=>`, `du=-sinxdx`

The integral is

`I=int(sinxdx)/(cos^5x)=-int(du)/u^5`

`=1/(4u^4)`

`=1/(4cos^4x)+C`