Silly u-substitution mistake happening somewhere

[Flammadeao]

Homework Statement

\int\frac{sin(2x)}{1+cos(x)^2}

Homework Equations

None?

The Attempt at a Solution

I know I can use a trig identity to end up with a numerator of -- 2sin(x)cos(x)

So:


\int\frac{2sin(x)cos(x)}{1+cos(x)^2}



I am using u=1+cos(x)^2 and du=-2sin(x)dx

Substitute in and I end up with

\int\frac{-cos(x)}{u}du

And that's where I hit a wall, because I still have a cos(x) in there. Anyone willing to offer hints on this one? Thanks much!

 

[Kurdt]

Try u=cos^2(x).

 

[Flammadeao]

Oh man, that would make sense... Gah, thankyou :)

 

[Elucidus]

I am using u=1+cos(x)^2 and du=-2sin(x)dx

If u=1+\cos^2(x) then du \neq -2\sin(x)\;dx

It would be du=-2\sin(x)\cos(x)\;dx

--Elucidus