# A trigonometric substitution problem

1. Nov 22, 2008

### celeramo

1. The problem statement, all variables and given/known data

$$\int$$sin3cos2xdx

3. 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 :)

2. Nov 22, 2008

### 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))

3. Nov 24, 2008

### celeramo

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