chebyshev polynomials are orthogonal

Let $⟨f,g⟩={\int }_{-1}^{1}f(x)g(x)\frac{1}{\sqrt{1-x^2}}dx$

The Chebyshev Polynomials $T_0,T_1,\dots$ are orthogonal with respect to the inner product $⟨\cdot ,\cdot ⟩$.

$$\begin{array}{rl}{\int }_{-1}^1{T}_n(x)T_m(x)\frac{1}{\sqrt{1-x^2}}dx& ⟹x=\cos \theta ,\frac{dx}{d\theta }=\sin \theta \\ & ={\int }_{\pi }^2\pi T_n(\cos \theta )T_m(\cos \theta )\frac{d\theta \sin \theta }{\sqrt{1-{\cos }^2\theta }}\\ & ={\int }_{\pi }^2\pi \cos (n\theta )\cos (m\theta )d\theta \text{ since }{T}_n=\cos (n\theta )\\ & ={\int }_{\pi }^2\pi \frac{e^{in\theta }+e^{-in\theta }}{2}+\frac{e^{im\theta }+e^{-im\theta }}{2}d\theta \\ & =\frac{1}{4}{\int }_{\pi }^{2\pi }{e}^{in\theta }+e^{-in\theta }+e^{im\theta }+e^{-im\theta }d\theta \\ & =\{\begin{array}{l}\frac{1}{4}{\int }_{\pi }^{2\pi }4d\theta =\theta {|}_{\pi }^{2\pi }=\pi \text{ if }n=m=0\\ \frac{1}{4}{\int }_{\pi }^{2\pi }2d\theta +\frac{1}{4}{\int }_{\pi }^{2\pi }2\cos ((n+m)\theta )d\theta =\frac{\pi }{2}+0=\frac{\pi }{2}\text{ if }n=m\ne 0\\ \frac{1}{4}{\int }_{\pi }^{2\pi }2\cos ((n+m)\theta )+2\cos ((n-m)\theta )d\theta =0\text{ if }n\ne m\end{array}\end{array}$$