A trigonometric substitution problem

[celeramo]

Homework Statement

\intsin³cos²xdx

The Attempt at a Solution

I've successfully solved this problem by factoring out 1 sinx and changing the sin²x to (1-cos²x then assigning u=cosx and du=-sinx and so on.

What I'm wondering is why does letting u=sin³x in the original integral not work. Then du=3cos²xdx and there is a cos²x in the original integral. Why does this fail? Please and thank you :)

 

[kittybobo1]

Well, the reason that doesn't work is that that is not the correct derivative. the derivative is 3sin(x)^2*cos(x) :) remember taking the derivative of v^3 is 3v^2*dv (where I let v=sin(x))

 

[celeramo]

Thanks very much. I don't know what I was thinking