Question

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

• A. $0$
• B. $1$
• C. $\sqrt{2}$
• D. $2\sqrt{2}$

Answer 1

The correct option is A $0$

$I={\int }_0^{\pi }\sqrt{1+\cos 2x}dx$

$={\int }_0^{\pi }\sqrt{2{\cos }^{2}x}dx$

$=\sqrt{2}{\int }_0^{\pi }\cos xdx$

$=\sqrt{2}{\left(sinx\right)}_0^{\pi }$

$=0$