Evaluate $\int_0^\pi \frac{\sin^2 nx}{\sin^2 x}dx$

Question

How do I evaluate this definite integral, I'm not even getting a slightest idea on how to approach this. Tried converting into cos using double angle property but that didn't help.

$$\int_0^\pi \frac{\sin^2 nx}{\sin^2 x}dx$$

Answer 1

Hint:

De Moivre's Formula: https://en.wikipedia.org/wiki/De_Moivre%27s_formula turns this integral into a rational function of sines and cosines.

Then you can use Weierstrass' tangent half-angle substitution: https://en.wikipedia.org/wiki/Tangent_half-angle_substitution.

Answer 2

$$\frac{\sin nx}{\sin x}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}} =\sum_{r=1}^ne^{i(n+1-2r)x}$$ $$\frac{\sin^2 nx}{\sin^2 x} =n+\sum_{(r,s)\in A}e^{i(2n+2-2r-2s)x} =n+\sum_{(r,s)\in A}\cos(2n+2-2r-2s)x $$ where $A$ is the set of $(r,s)$ with $r$, $s\in\{1,2,\ldots,n\}$ with $r+s\ne n+1$. The details of this are not important as $\int_0^\pi \cos mx\,dx=0$ for nonzero integers $m$. Therefore $$\int_0^\pi\frac{\sin^2 nx}{\sin^2 x}\,dx=n\pi.$$

Answer 3

Let $$I_{n}=\int_{0}^{\pi}\frac{\sin^2nx}{\sin^2x} dx$$

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

$$I_{n+1}-2I_{n}+I_{n-1}=0$$

Thus $I_{1},I_{2},I_{3},...$ are in A.P

$I_{1}=\pi$

$I_{2}=2\pi$

So, $I_{n}=n\pi$