Question

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

Answer 1

$$\begin{gathered} I={\int }_0^{2\mathrm{\pi }}\sqrt{1+\sin \frac{x}{2}}dx \\ ={\int }_0^{2\mathrm{\pi }}\sqrt{{\cos }^2\frac{x}{4}+{\sin }^2\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}}dx \\ ={\int }_0^{2\mathrm{\pi }}\sqrt{{\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)}^2}dx \\ ={\int }_0^{2\mathrm{\pi }}\left|\cos \frac{x}{4}+\sin \frac{x}{4}\right|dx \end{gathered}$$

When $0\le x\le 2\mathrm{\pi }$, $0\le \frac{x}{4}\le \frac{\mathrm{\pi }}{2}$
$$\begin{gathered} \therefore \sin \frac{x}{4}\ge 0,\cos \frac{x}{4}\ge 0 \\ \Rightarrow \cos \frac{x}{4}+\sin \frac{x}{4}\ge 0 \\ \Rightarrow \left|\cos \frac{x}{4}+\sin \frac{x}{4}\right|=\cos \frac{x}{4}+\sin \frac{x}{4} \end{gathered}$$

$$\begin{gathered} \therefore I={\int }_0^{2\mathrm{\pi }}\left(\cos \frac{x}{4}+\sin \frac{x}{4}\right)dx \\ ={\overline{)\frac{\sin {\displaystyle \frac{x}{4}}}{{\displaystyle \frac{1}{4}}}}}_0^{2\mathrm{\pi }}+{\overline{)\frac{\left(-\cos {\displaystyle \frac{x}{4}}\right)}{{\displaystyle \frac{1}{4}}}}}_0^{2\mathrm{\pi }} \\ =4\left(\sin \frac{\mathrm{\pi }}{2}-\sin 0\right)-4\left(\cos \frac{\mathrm{\pi }}{2}-\cos 0\right) \\ =4\left(1-0\right)-4\left(0-1\right) \\ =4+4 \\ =8 \end{gathered}$$