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