A calculus problem by Ravneet Singh

Let $\displaystyle I(n) = \int_{0}^{\frac{\pi}{2}} \frac{\sin ((2n-1)x)}{\sin x}dx$ for all natural numbers $n$ .

$\displaystyle I(n+1) - I(n) = \int_{0}^{\frac{\pi}{2}} \frac{\sin ((2n+1)x) - \sin ((2n-1)x)}{\sin x}dx = \int_{0}^{\frac{\pi}{2}} \frac{2\cos{2nx}\sin x}{\sin x} dx$

$\displaystyle = \int_{0}^{\frac{\pi}{2}} 2\cos{2nx} = \frac{\sin{2n\frac{\pi}{2}}}{n} - \frac{\sin{2n \cdot 0}}{n} = 0$ .

It follows that $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin (2017x)}{\sin x}dx = I(1009) = I(1008) = .... = I(1) = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x}dx = \frac{\pi}{2}$