Although your problem is classic, in order to prove the identity, we need to introduce Chebyshev's polynomials of the second kind:
$$\begin{cases}
U_0 (x) = 1\\
U_n (x) = \frac{\sin \left[(n+1)\arccos x\right]}{\sin (\arccos x)}, n\geq 1.
\end{cases}$$
Now the below integral must hold for all $n$:
$$I_n:=\int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t = \pi \cdot T_n (x)$$
To prove the integral stated above by using the recursion formula of $T_n (x)$ and $U_n (x)$ for induction. Hence it is vital to prove it's true for $n=1,2$. Initially, we have:
$$\int_{0}^{\pi} \frac{1}{\cos x - \cos v}dx = \frac{1}{2}\int_{-\pi}^{\pi}\frac{1}{\cos x - \cos v}dx = \frac{1}{2}\int_{-\pi}^{\pi} \frac{e^{ix}}{(e^{ix} - e^{iv})(e^{ix}-e^{-iv})}~dx$$$$=\int_{\vert z \vert= 1}\frac{-i}{(z - e^{iv})(z-e^{-iv})}~dz = -\frac{i}{e^{iv}-e^{-iv}}\left(\int_{\vert z\vert=1}\frac{1}{z-e^{iv}}dz-\int_{\vert z\vert=1}\frac{1}{z-e^{-iv}}dz\right)=0$$
For $n=0$:
$$I_1= \int_{-1}^{1}\frac{\sqrt{1-t^2}}{\cos \theta-t}~\mathrm{d}t=\int_{-\pi}^{\pi}\frac{\sin^2 k}{2(\cos \theta-\cos k)}~\mathrm{d}k=\int_{-\pi}^{\pi}\frac{1-\cos 2k}{4(\cos \theta-\cos k)}~\mathrm{d}k$$$$=\int_{-\pi}^{\pi}\frac{\cos 2k}{4(\cos k-\cos \theta )}~\mathrm{d}k$$
The last expression of $I_1$ is worth trying to train complex analysis method. Then we obtain: $I_1 = \pi x = \pi\cdot T_1 (x)$. The second integral follows easily since:
$$I_2 = \int_{-1}^{1} \frac{2t\cdot\sqrt{1-t^2}}{x-t}~\mathrm{d}t= \int_{-1}^{1} \frac{2x\cdot\sqrt{1-t^2}}{x-t}~\mathrm{d}t -\int_{-1}^{1} 2\cdot\sqrt{1-t^2}~\mathrm{d}t$$
$$=2x\cdot I_1 - \pi = \pi\cdot(2x^2-1)=\pi T_1(x)$$
At this time, we will assume $I_k = \pi \cdot T_k(x)$ from $k=1,...,n$. The work to do now is to show the equality holds for $k=n+1$.
$$I_{n+1} = \int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n}(t)}{x-t}~\mathrm{d}t= \int_{-1}^{1}\frac{2t\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t- \int_{-1}^{1}\frac{\sqrt{1-t^2}\cdot U_{n-2}(t)}{x-t}~\mathrm{d}t$$
$$= \int_{-1}^{1}\frac{2x\sqrt{1-t^2}\cdot U_{n-1}(t)}{x-t}~\mathrm{d}t - 2\int_{-1}^{1} \sqrt{1-t^2}U_{n-1}(t)dt - \pi \cdot T_{n-1} (x)= \pi \cdot 2x\cdot T_n (x) - \pi \cdot T_{n-1} (x)$$$$\Rightarrow I_{n+1}= \pi \cdot T_{n+1} (x)$$
In the calculation above, we have:
$$\int_{-1}^{1} U_{n-1} (x)\sqrt{1-x^2}\mathrm{d}x = \int_{0}^{\pi} \sin t\cdot \sin nt dt = \frac{1}{2}\left(\int_{0}^{\pi} \cos (n-1)t \mathrm{d}t - \int_{0}^{\pi} \cos (n+1)t\mathrm{d}t \right)=0 $$
For the integral $I_n$, substitute $t = \cos u$ and integrate by parts:
$$I_n = \int_{0}^{\pi} \frac{\sin^2 u\cdot U_{n-1} (\cos u)}{x-\cos u} ~ \mathrm{d}u= \int_{0}^{\pi} \frac{\sin t\cdot \sin nt}{x-\cos t}~\mathrm{d}t$$$$=\ln \vert x - \cos u\vert. \sin nt\Bigg|_0^{\pi}-n\cdot\int_{0}^{\pi}\cos nu.\ln \vert x - \cos u\vert\mathrm{d}u=-n\cdot f_n(x) $$
$$\Rightarrow f_n (x) = \frac{\pi}{n}\cdot T_n(x)$$
$n=m=0 \Rightarrow \int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\int \frac{dx}{\sqrt{1-x^2}}=arcsin x$
$\int \frac{dx}{\sqrt{1-x^2}}=arcsin x \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=\pi$
$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 $m\neq n$
$x=cos\theta \Rightarrow dx=-sin\theta d\theta$
$x=cos\theta \Rightarrow T_m(x)=\cos m\theta$
$\int T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=-\int \cos m\theta \cos n\theta{d\theta}$
$\int \cos m\theta \cos n\theta{d\theta}=\frac{\sin(m-n)\theta}{2(m-n)}+\frac{\sin(m+n)\theta}{2(m+n)} \Rightarrow \int_{-1}^{1} T_{n}(x)\,T_{m}(x)\,{\frac {\mathrm {d} x}{\sqrt {1-x^{2}}}}=0$
Best Answer
It is actually not that hard. You can derive a lot of relations on the wiki page yourself by substituting $\cos\theta$ for $x$ and use the defining relation of Chebyshev polynomials:
$$T_n(\cos\theta) = \cos( n\theta)$$
For example, one have: $$\begin{align}\int T_n(x) dx = & \int T_n(\cos\theta) d\cos \theta\\ = & -\int \cos(n\theta)\sin\theta d\theta\\ = & -\frac12 \int \left(\sin((n+1)\theta) - \sin((n-1)\theta)\right)d\theta\\ = & \frac12 \left(\frac{\cos((n+1)\theta)}{n+1} - \frac{\cos((n-1)\theta)}{n-1}\right) + \text{const.}\\ = & \frac12 \left(\frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1}\right) + \text{const.} \end{align} $$