Question

Evaluate : ${\int }_{-\pi /2}^{199\pi /2}\sqrt{1+\cos 2x}dx$ is

Answer 1

${\int }_{-\frac{\pi }{2}}^{199\frac{\pi }{2}}\sqrt{1+cos2x}{\int }_{-\frac{\pi }{2}}^{199\frac{\pi }{2}}\sqrt{1+2{\cos }^{2}x-1}=\sqrt{2}{\int }_{-\frac{\pi }{2}}^{199\frac{\pi }{2}}\left|cosx\right|dx=199\sqrt{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos xdx=199\sqrt{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sin x=199\sqrt{2}(1-(-1\left)\right)=298\sqrt{2}$