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

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    $\begingroup$ $x=\cos\theta$ means $dx=-\sin\theta\,d\theta$, wouldn't you think so? $\endgroup$
    – qfwfq
    Mar 6 '21 at 7:01
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    $\begingroup$ That's why you don't just omit dx in integrals. $\endgroup$ Mar 6 '21 at 8:08
  • $\begingroup$ I'm sorry! You are right. I had forgotten to write $dx$ $\endgroup$
    – Mark
    Mar 7 '21 at 7:15
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$n=m=0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int \frac{dx}{\sqrt{1-x^2}}=arcsin x$

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

$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 $m\neq n$

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

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

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

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

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