Question

$\int \frac{x+\sin x}{1+\cos x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{x+\sin x}{1+\cos x}\right)dx \\ =\int \left[\frac{x}{1+\cos x}+\frac{\sin x}{1+\cos x}\right]dx \\ =\int \left[\frac{x}{2{\cos }^2{\displaystyle \frac{x}{2}}}+\frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}\right]dx \\ =\frac{1}{2}\int \underset{\mathrm{I}}{x}·\underset{\mathrm{II}}{{\sec }^2}\frac{x}{2}dx+\int \tan \frac{x}{2}dx \\ =\frac{1}{2}\left[x·\frac{\tan \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{1}{2}}}-\int 1\times 2\tan \left(\frac{x}{2}\right)dx\right]+\frac{\log \left|sec{\displaystyle \frac{x}{2}}\right|}{{\displaystyle \frac{1}{2}}}+C \\ =x\tan \left(\frac{x}{2}\right)-\frac{\log \left|\sec {\displaystyle \frac{x}{2}}\right|}{{\displaystyle \frac{1}{2}}}+\log \frac{\left|\sec {\displaystyle \frac{x}{2}}\right|}{{\displaystyle \frac{1}{2}}}+C \\ =x\tan \left(\frac{x}{2}\right)+C \end{gathered}$$