- #1

- 9

- 0

## Homework Statement

[tex]\int[/tex]sin

^{3}cos

^{2}xdx

## The Attempt at a Solution

I've successfully solved this problem by factoring out 1 sinx and changing the sin

^{2}x to (1-cos

^{2}x then assigning u=cosx and du=-sinx and so on.

What I'm wondering is why does letting u=sin

^{3}x in the original integral not work. Then du=3cos

^{2}xdx and there is a cos

^{2}x in the original integral. Why does this fail? Please and thank you :)