Question

What is ${\int }_0^{2\pi }\sqrt{1+\sin \frac{x}{2}}dx$ equal to?

• A. $8$
• B. $4$
• C. $2$
• D. $0$

Answer 1

The correct option is A $8$

$1+\sin \frac{x}{2}={\sin }^2\frac{x}{4}+{\cos }^2\frac{x}{4}+2\sin \frac{x}{4}\cos \frac{x}{4}{(\sin \frac{x}{4}+\cos \frac{x}{4})}^2\sqrt{1+\sin \frac{x}{2}}=\sqrt{{(\sin \frac{x}{4}+\cos \frac{x}{4})}^2}=\sin \frac{x}{4}+\cos \frac{x}{4}$

${\int }_0^{2\pi }(\sin \frac{x}{4}+\cos \frac{x}{4})dx=4(-\cos \frac{x}{4}+\sin \frac{x}{4})+cAfterputtingthelimitweget\Rightarrow 4(1-(-1\left)\right)=8$

So correct answer will be option A