Integral of $\int \frac{1+\sin(2x)}{\operatorname{tg}(2x)}dx$ I'm trying to find the $F(x)$ of this function but I don't find how to do it, I need some hints about the solution. 
I know that $\sin(2x) = 2\sin(x)\cos(x)$ its help me? It's good way to set $2x$ as $t$?
$$\int \frac{1+\sin(2x)}{\operatorname{tg}(2x)}dx$$
EDIT

Its right to do it like that? 
$$\int \frac{1+2\sin(x)\cos(x)dx}{\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}}$$
$$\int \left(\frac{\cos^2(x)-\sin^2(x)}{2\sin(x)\cos(x)}+\cos^2(x)-\sin^2(x)\right)dx = \int \left(\operatorname{ctg}(2x)+\frac{1+\cos(2x)}{2}-\frac{1-\cos(2x)}{2}\right)dx$$ and the integral of $\displaystyle \operatorname{ctg}(2x) = \frac{\ln|\sin(2x)|}{2}+C$ and the two other is equal to $\displaystyle \frac{\sin(2x)}{2}+C$
Thanks

Hint: Set $y={\rm tg} x$ and express all the functions through $y$.
