Question

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

• A. 2
• B. 4
• C. 8

Answer 1

The correct option is C 8

The given integral is ${\int }_0^{2\pi }\sqrt{1+\sin \frac{x}{2}}dx$
Now, ${\sin }^2\frac{x}{4}+{\cos }^2\frac{x}{4}=1$
And, $2\sin \frac{x}{4}\cos \frac{x}{4}=\sin (2\times \frac{x}{4})$
Thus, $\sqrt{1+\sin \frac{x}{2}}=\sqrt{{\sin }^2\frac{x}{4}+{\cos }^2\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}}$
$=\sqrt{{(\sin \frac{x}{4}+\cos \frac{x}{4})}^2}$
$=|\sin \frac{x}{4}+\cos \frac{x}{4}|$

Hence, ${\int }_0^{2\pi }\sqrt{1+\sin \frac{x}{2}}dx=={\int }_0^{2\pi }|sin\frac{x}{4}+cos\frac{x}{4}|dx=4{[sin\frac{x}{4}-cos\frac{x}{4}]}_0^{2\pi }$
$=4[1-0-0+1]=8$