Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{x+\sin x}{1+\cos x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I=\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\frac{x+\sin x}{1+\cos x}dx \\ ={\int }_0^{\mathrm{\pi }/2}\frac{x+\sin x}{2{\cos }^2{\displaystyle \frac{x}{2}}}dx \\ ={\int }_0^{\mathrm{\pi }/2}\left[\frac{x}{2{\cos }^2{\displaystyle \frac{x}{2}}}+\frac{\sin x}{2{\cos }^2{\displaystyle \frac{x}{2}}}\right]dx \\ =\frac{1}{2}{\int }_0^{\mathrm{\pi }/2}xse{c}^2\frac{x}{2}dx+{\int }_0^{\mathrm{\pi }/2}\frac{2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}}dx \\ =\frac{1}{2}{\left[x\frac{\tan {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{1}{2}}}\right]}_0^{\mathrm{\pi }/2}-\frac{1}{2}{\int }_0^{\mathrm{\pi }/2}\frac{\tan {\displaystyle \frac{x}{2}}}{{\displaystyle \frac{1}{2}}}dx+{\int }_0^{\mathrm{\pi }/2}\tan \frac{x}{2}dx \\ ={\left[x\tan \frac{x}{2}\right]}_0^{\mathrm{\pi }/2}-{\int }_0^{\mathrm{\pi }/2}\tan \frac{x}{2}dx+{\int }_0^{\mathrm{\pi }/2}\tan \frac{x}{2}dx \\ =\left[\frac{\mathrm{\pi }}{2}\tan \frac{\mathrm{\pi }}{4}\right] \\ =\frac{\mathrm{\pi }}{2}\times 1 \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$