Prove the orthogonality relation of Chebyshev polynomials of the first kind

Question

The Chebyshev polynomials of the first kind are obtained from the recurrence relation
$$\begin{aligned}T_{0}(x)&=1\\T_{1}(x)&=x\\T_{n+1}(x)&=2x\,T_{n}(x)-T_{n-1}(x)~.\end{aligned}$$
Prove that:
$$\int _{-1}^{1}T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}={\begin{cases}0&~{\text{ if }}~n\neq m~,\\\\\pi &~{\text{ if }}~n=m=0~,\\\\{\frac {\pi }{2}}&~{\text{ if }}~n=m\neq 0~.\end{cases}}$$
I tried to prove it using $x = \cos \theta$ and using the defining identity $T_n(\cos \theta) = \cos n\theta$, but I couldn't… I arrived to this integral, which for me is hard to solve:
$$2\int_0^{\pi/2}\frac{\cos n\theta\, \cos m\theta}{\sin \theta}$$

Answer 1

$n=m=0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int \frac{dx}{\sqrt{1-x^2}}=arcsin x$

$\int \frac{dx}{\sqrt{1-x^2}}=arcsin x \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\pi$

$x=cos \theta $ and $n=m\neq 0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=-\int\cos^2n\theta d\theta$

$-\int\cos^2n\theta d\theta=-(\frac \theta2+\frac{\sin 2n\theta}{4n}) \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\frac\pi 2 $

let $m\neq n$

$x=cos\theta \Rightarrow dx=-sin\theta d\theta$

$x=cos\theta \Rightarrow T_m(x)=\cos m\theta$

$\int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=-\int \cos m\theta \cos n\theta{d\theta}$

$\int \cos m\theta \cos n\theta{d\theta}=\frac{\sin(m-n)\theta}{2(m-n)}+\frac{\sin(m+n)\theta}{2(m+n)} \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=0$