Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$ 
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$

My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?

From where you are you did a great job, note that


$$(\tan (\frac{x}{2}))' =\frac{1}{2}(1+\tan^2 (\frac{x}{2}))~~~and~~~ -2(\ln|\cos  (\frac{x}{2})|)' = \tan (\frac{x}{2})$$
Therefore $$\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx =  \dfrac {1}{2}\int (\tan^2 (\dfrac {x}{2}) +1) dx + \int \tan (\dfrac {x}{2}) dx  \\=\tan (\frac{x}{2}) -2\ln|\cos  (\frac{x}{2})|+c$$
