Question

$${\int }_0^{2\pi }\sqrt{1+\sin \frac{x}{2}}dx$$

Answer 1

$$I={\int }_0^{2\pi }\sqrt{1+\sin \frac{x}{2}}dx$$
$$={\int }_0^{2\pi }\sqrt{{\cos }^2\frac{x}{4}+{\sin }^2\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}}dx$$
$$={\int }_0^{2\pi }\sqrt{{(\cos \frac{x}{4}+\sin \frac{x}{4})}^2}dx$$
$$={\int }_0^{2\pi }|\cos \frac{x}{4}+\sin \frac{x}{4}|dx$$

When

$$0\le x\le 2\pi$$

$$0\le \frac{x}{4}\le \frac{\pi }{2}$$

$$\therefore \sin \frac{x}{4}\ge 0,\cos \frac{x}{4}\ge 0$$
$$\Rightarrow \cos \frac{x}{4}+\sin \frac{x}{4}\ge 0$$
$$\Rightarrow |\cos \frac{x}{4}+\sin \frac{x}{4}|=\cos \frac{x}{4}+\sin \frac{x}{4}$$

$$\therefore I={\int }_0^{2\pi }(\cos \frac{x}{4}+\sin \frac{x}{4})dx$$
$$={\frac{\sin \frac{x}{4}}{\frac{1}{4}}|}_0^{2\pi }+{\frac{(-\cos \frac{x}{4})}{\frac{1}{4}}|}_0^{2\pi }$$
$$=4(\sin \frac{\pi }{2}-\sin 0)-4(\cos \frac{\pi }{2}-\cos 0)$$
$$=4(1-0)-4(0-1)$$
$$=4+4$$
$$=8$$