Question

Find m if :${\int }_0^{\pi }\left(\frac{\sqrt{1+\cos 2x}}{2}\right)dx=\sqrt{m}$

Answer 1

${\int }_0^{\pi }\left(\frac{\sqrt{1+\cos 2x}}{2}\right)dx$

$={\int }_0^{\pi }\left(\frac{\sqrt{1+2{\cos }^{2}x-1}}{2}\right)dx$

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

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

$=\frac{\sqrt{2}}{2}[\sin \pi -\sin 0]=0$

$\therefore m=0$