Prove that, $\int_0^{\pi/2}\frac{\sin (2n+1)x}{\sin x}~dx=\frac{\pi}{2}$ Prove that, $$\int_0^{\pi/2}\dfrac{\sin (2n+1)x}{\sin x}~dx=\dfrac{\pi}{2}$$  For every integer $n\ge 0$
I tried using induction without any help. Possibly my method of application was not right. Please help me!

Firstly, note that $$\sin (z+2)x - \sin zx = 2\sin x \cos (z+1) x \iff \frac{\sin (z+2)x - \sin zx}{\sin x} = 2\cos (z+1)x$$
Set $z=2n-1$ and integrate to get:
$$\int_0^{\frac{\pi}{2}} \frac{\sin (2n+1)x}{\sin x} dx - \int_0^{\frac{\pi}{2}} \frac{\sin (2n-1)x}{\sin x}dx = \int_0^{\frac{\pi}{2}} 2\cos (2n)x\ dx = \frac{2}{2n} \sin \left[(2n)\cdot \frac{\pi}{2}\right] = 0$$
Therefore we get the recurrence relation:
$$\int_0^{\frac{\pi}{2}} \frac{\sin (2n+1)x}{\sin x} dx = \int_0^{\frac{\pi}{2}} \frac{\sin (2n-1)x}{\sin x}dx =\cdots = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x}dx = \frac{\pi}{2}$$
