Question

The value of $\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{\sin (2m+1)x}{\sin x}dx$ is

Answer 1

We know
$2\sin x[\cos x+\cos 3x+\cos 5x+...+\cos (2k-1)x]=2\sin x\cos x+2\sin x\cos 3x+2\sin x\cos 5x+...+2\sin x\cos (2k-1)x=\sin 2x+(\sin 4x-\sin 2x)+(\sin 6x-\sin 4x)+...+(\sin 2kx-\sin (2k-2)x)=\sin 2kx$
$\therefore 2[\cos x+\cos 3x+\cos 5x+...+\cos (2k-1)x]=\frac{\sin 2kx}{\sin x}$
Now $\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{\sin (2m+1)x}{\sin x}dx$
$=\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\sin 2mx\cot xdx+\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\cos (2m+1)xdx$
$=\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}dx+\frac{8}{\pi }{\int }_0^{\frac{\pi }{2}}[\cos x+\cos 3x+\cos 5x+...+\cos (2k-1)x]dx$
$+\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\cos \left(2mx\right)dx+\frac{4}{\pi }{\int }_0^{\frac{\pi }{2}}\cos (2m+1)xdx$
$=\frac{4}{\pi }\left(\frac{\pi }{2}\right)=2$