Question

The value of $\underset{0}{\overset{2\mathrm{\pi }}{\int }}\sqrt{1+\sin \frac{x}{2}}dx$ is
(a) 0
(b) 2
(c) 8
(d) 4

Answer 1

(c) 8

$$\begin{gathered} {\int }_0^{2\mathrm{\pi }}\sqrt{1+\sin \frac{x}{2}}dx \\ ={\int }_0^{2\mathrm{\pi }}\sqrt{{\sin }^2\frac{x}{4}+{\cos }^2\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}}dx \\ ={\int }_0^{2\mathrm{\pi }}\left(\sin \frac{x}{4}+\cos \frac{x}{4}\right)dx \\ ={\left[\frac{-\cos \frac{x}{4}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}+\frac{\sin \frac{x}{4}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\right]}_0^{2\mathrm{\pi }} \\ =4{\left[\sin \frac{x}{4}-\cos \frac{x}{4}\right]}_0^{2\mathrm{\pi }} \\ =4\left[\sin \frac{2\mathrm{\pi }}{4}-\cos \frac{2\mathrm{\pi }}{4}-\sin 0+\cos 0\right] \\ =4\left[\sin \frac{\mathrm{\pi }}{2}-\cos \frac{\mathrm{\pi }}{2}-0+1\right] \\ =4\left[1-0-0+1\right] \\ =4\times 2 \\ =8 \end{gathered}$$