Reduction formulae in definite integration

Question

$$I_n = \int_0^{\pi}\frac{\sin^2(nx)}{\sin^2(x)}dx $$ Find relation between $I_n$, $I_{n+1}$ and $I_{n+2}$ I tried integration by parts by taking $\sin^2(nx)$ as the first function, but reached nowhere.

Answer 1

Using Chebyshev polynomials of the second kind,

$$ I_n = \int_{0}^{\pi} U_{n-1}(\cos x)^2\,dx = \int_{-1}^{1} \frac{U_{n-1}^2(z)}{\sqrt{1-z^2}}\,dz = \color{red}{\pi n}\tag{1} $$ since Chebyshev polynomials of the second kind are an orthogonal base of $L^2(-1,1)$ with the inner product $\langle f,g\rangle = \int_{-1}^{1}\frac{f(x)g(x)}{\sqrt{1-x^2}}\,dx $. That also follows from: $$ \frac{\sin(nx)}{\sin(x)}=\frac{e^{nix}-e^{-nix}}{e^{ix}-e^{-ix}}=\left\{\begin{array}{rcl}1+2\left(\cos((n-1)x)+\cos((n-3)x)+\ldots+\cos(2x)\right)&\text{if}&n\equiv{1}\pmod{2}\\2\left(\cos((n-1)x)+\cos((n-3)x)+\ldots+\cos(x)\right)&\text{if}&n\equiv{0}\pmod{2}\end{array}\right.$$ and the usual orthogonality properties of the sine and cosine functions, as a base of $L^2(-\pi,\pi)$ with the usual inner product.

Answer 2

Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,

$$\sin^2(n+1)x-\sin^2nx=\sin x\cdot\sin(2n+1)x$$

$$I_{n+1}-I_n=\int_0^\pi\dfrac{\sin(2n+1)}{\sin x}\ dx=J_n\text{(say)}$$

Now using Prosthaphaeresis Formulas, $$\sin(2m+1)x-\sin\{2(m-1)+1\}x=2\sin x\cos(2m)x$$

$$\implies J_m-J_{m-1}=\int_0^\pi2\cos2mx\ dx=\cdots=0$$

$$\implies J_{m+1}-J_m=0\iff J_{m+1}=J_m$$

Now $$J_0=\int_0^\pi\dfrac{\sin(2\cdot0+1)}{\sin x}\ dx=\cdots=\pi-0$$

Answer 3

$$I_{n+1}-I_n = \int_0^{\pi}\frac{\sin^2(n+1)x}{\sin^2x}-\frac{\sin^2nx}{\sin^2x}dx$$ $$=\int_0^\pi \frac{\sin (2n+1)x}{\sin x}dx=\pi$$ This means that the values of these integrals form an AP, and we know that $I_1=\pi$. Thus we conclude that the value of $I_n=n\pi$.