Prove the orthogonality relation of Chebyshev polynomials of the first kind

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: $$2\int_0^{\pi/2}\frac{\cos n\theta\, \cos m\theta}{\sin \theta}$$

• $x=\cos\theta$ means $dx=-\sin\theta\,d\theta$, wouldn't you think so? Mar 6 '21 at 7:01
• That's why you don't just omit dx in integrals. Mar 6 '21 at 8:08
• I'm sorry! You are right. I had forgotten to write $dx$
– Mark
Mar 7 '21 at 7:15

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

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

$$x=cos \theta$$ 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 $$m\neq n$$

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

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

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

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