Question

The value of the integral $\int\limits_0^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$ is equal to

Answer 1

$ I=60 \int\limits_0^{\pi / 2}\left(\frac{\sin 6 x-\sin 4 x}{\sin x}+\frac{\sin 4 x-\sin 2 x}{\sin x}+\frac{\sin 2 x}{\sin x}\right) d x$
$ I=60 \int\limits_0^{\pi / 2}(2 \cos 5 x+2 \cos 3 x+2 \cos x) d x$
$I=\left.60\left(\frac{2}{5} \sin 5 x+\frac{2}{3} \sin 3 x+2 \sin x\right)\right|_0 ^{\pi / 2}=104$