Question

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

• A. $0$
• B. $2$
• C. $1$
• D. $-1$

Answer 1

The correct option is B $2$

Let $I={\int }_0^{\pi }\sqrt{\frac{1+\cos 2x}{2}}dx={\int }_0^{\pi }\left|\cos x\right|dx$
$={\int }_0^{\pi /2}\cos xdx-{\int }_{\pi /2}^{\pi }\cos xdx$
$={\left[\sin x\right]}_0^{\pi /2}-{\left[\sin x\right]}_{\pi /2}^{\pi }$
$=[\sin \frac{\pi }{2}-\sin 0]-[\sin \pi -\sin \frac{\pi }{2}]$
$=1+1=2$