Question

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

Answer 1

$$LetI=\underset{0}{\overset{\pi /2}{\int }}\frac{x+\sin x}{1+\cos x}dx$$
$$={\int }_0^{\pi /2}\frac{x+\sin x}{2{\cos }^2\frac{x}{2}}dx$$
$$={\int }_0^{\pi /2}[\frac{x}{2{\cos }^2\frac{x}{2}}+\frac{\sin x}{2{\cos }^2\frac{x}{2}}]dx$$
$$=\frac{1}{2}{\int }_0^{\pi /2}xse{c}^2\frac{x}{2}dx+{\int }_0^{\pi /2}\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}}dx$$
$$=\frac{1}{2}{\left[x\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]}_0^{\pi /2}-\frac{1}{2}{\int }_0^{\pi /2}\frac{\tan \frac{x}{2}}{\frac{1}{2}}dx+{\int }_0^{\pi /2}\tan \frac{x}{2}dx$$
$$={[x\tan \frac{x}{2}]}_0^{\pi /2}-{\int }_0^{\pi /2}\tan \frac{x}{2}dx+{\int }_0^{\pi /2}\tan \frac{x}{2}dx$$
$$=[\frac{\pi }{2}\tan \frac{\pi }{4}]$$
$$=\frac{\pi }{2}\times 1$$
$$=\frac{\pi }{2}$$