trigonometric function integration I have this integral: $$ \int \dfrac{\sin(2x)}{1+\sin^2(x)}\,dx$$ and I need hint solving it. I tried using the trigonometric identities and let $$u=\sin(x)$$ but I got $$\int ... =\int \dfrac{2u}{1+u^2}\, du$$ which I don't know how to solve. I also tried letting $$u=tg\left(\frac{x}{2}\right)$$ but that leads to $$\int ...=\int \frac{8t(1-t^2)}{(1+t^2)(t^4+6t^2+1)} du$$ which again I can't solve. I'll be glad for help.

Hint: $\displaystyle  \log (u(x))'=\frac{u'(x)}{u(x)}$, (with the implied assymption that $u(x)>0$ for all $x$ in the domain of $u$).
You should note, however, that $\displaystyle  \log (|u(x)|)'=\frac{u'(x)}{u(x)}$, for all $x\in \operatorname{dom}(u)$ such that $u(x)\neq 0$. Also $1+u^2>0$.
