Question

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

• A. $4$
• B. $6$
• C. $8$
• D. None of these

Answer 1

The correct option is A $4$

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

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

$={\int }_0^{2\pi }\sqrt{{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)}^2}dx$

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

$={\left[-2\cos \frac{x}{2}+2\sin \frac{x}{2}\right]}_0^{2\pi }$

$=2{\left[\sin \frac{x}{2}-\cos \frac{x}{2}\right]}_0^{2\pi }$

$=2\left[1+1\right]$

$=4$