∫ 0 π / 2 x + sin x 1 + cos x d x =

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x+\sin x}{1+\cos x}. dx$

($1+\cos x=2{\cos }^2\frac{x}{2}$ as $\cos 2x=2{\cos }^{2}x-1$)

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{x+\sin x}{2{\cos }^2{\displaystyle \frac{x}{2}}}. dx$

$\left(\sin 2x=2\sin x \cos x \& \frac{1}{\cos x}=\sec x\right)$

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{x{\sec }^2{\displaystyle \frac{x}{2}}}{2}+\frac{\cancel{2}\sin {\displaystyle \frac{x}{2}}\cancel{\cos }{\displaystyle \frac{x}{2}}}{\cancel{2} {\cos }^2{\displaystyle \frac{x}{2}}}\right).dx$

$I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{1}{2}\left(x.{\sec }^2\frac{x}{2}\right)dx +{\int }_0^{\frac{\pi }{2}}\tan \frac{x}{2}.dx$

Using chain rule intergration, we get

$I=\frac{1}{2}\left(\right\{x{\int }_0^{\frac{\pi }{2}} {\sec }^2\frac{x}{2}. dx-{\int }_0^{\frac{\pi }{2}}\left(\frac{d}{dx}\left(x\right)\int {\sec }^2\frac{x}{2}.dx\right)+{\int }_0^{\frac{\pi }{2}}\tan \frac{x}{2}dx$

$I=\frac{1}{2}\left(\{x 2\tan \frac{x}{2}\right){\int }_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}2 \tan \left(\frac{x}{2}\right)dx+ {\int }_0^{\frac{\pi }{2}}\tan \frac{x}{2}dx$

$I={\left(x\tan \frac{x}{2}\right)}_0^{\frac{\pi }{2}}{\int }_0^{\frac{\pi }{2}} \cancel{\tan }\left(\frac{x}{2}\right)dx+{\int }_0^{\frac{\pi }{2}}\cancel{\tan }\frac{x}{2}dx$

$I=\left(\frac{\mathrm{\pi }}{2} \tan \left(\frac{\mathrm{\pi }}{4}\right)-0\right)=\frac{\mathrm{\pi }}{2}$.