Integrating Ratio Of Sines?

Claim : $\displaystyle \int_0^\pi \dfrac{\sin^2 nx}{\sin^2 x} \, dx = n \pi$ , where $n$ is a positive integer .

Proof : Let $\displaystyle I_n = \int_0^\pi \dfrac{\sin^2 nx}{\sin^2 x} \, dx$ , where $n = 1,2,3,\ldots$ .

It is trivial to show that $I_1 = \pi$ .

To prove that $I_n = n\pi$ is equivalent to proving that $I_{n+1} - I_n = (n+1)\pi - n\pi = \pi$ . We have

$\begin{aligned} I_{n+1} - I_n &=& \int_0^\pi \dfrac{ \sin^2 [ (n+1) x ] - \sin^2 (nx) }{\sin^2 x }\, dx\\ &=& \int_0^\pi \dfrac{ \left( \sin [ (n+1) x ] + \sin[ nx] \right) \left( \sin [ (n+1) x ] - \sin[ nx] \right) }{\sin^2 x} \, dx\\ &=& \int_0^\pi \dfrac{ 2\sin \left[ \frac{2n+1}2 x \right]\cos \frac x2 \cdot 2\cos \left[ \frac{2n+1}2 x \right ] \sin \frac x2 }{\sin^2 x} \, dx\\ &=& \int_0^\pi \dfrac{\sin[(2n+1)x] \cdot \sin x}{\sin^2 x} \, dx \\ &=& \int_0^\pi \dfrac{\sin[(2n+1)x] }{\sin x} \, dx \\ &=& \int_0^\pi \left [1 + 2\sum_{k=1}^n \cos(2kx) \right ] \, dx \\ &=& \pi + 2 \cancelto0{ \left [\sum_{k=1}^n \dfrac1{2k} \sin(2kx) \right]_0^\pi \quad } = \pi \; .\\ \end{aligned}$

Since the finite difference between each consecutive $I_n$ 's is $\pi$ , and we know that $I_1 = \pi$ , then $I_n= n\pi$ . We're done!

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In the steps above, we have used the trigonometric identities:

• Sum-to-product identities : $\sin A + \sin B = 2\sin \left( \dfrac{A+B}2 \right) \cos \left( \dfrac{A-B}2 \right)$ and $\sin A - \sin B = 2\cos \left( \dfrac{A+B}2 \right) \sin \left( \dfrac{A-B}2 \right)$ .

• Double angle identities : $\sin(2A) = 2\sin A \cos A$ .

• Dirichlet kernel, $\displaystyle \dfrac{\sin[(2n+1)x]}{\sin x} = 1 + 2\sum_{k=1}^n \cos(kx)$ , which can be shown using straightforward telescoping sum of product-to-sum identities.

I know it's cheating, but just let $n=1$ and you are done :D