Chebyshev polynomials of first kind I know the chebyshev polynomials of the first kind can be approximated using the cosine function, where $T_n(\cos \theta)=\cos(n \theta)$ and I know that chebyshev polynomials are a family of orthogonal polynomials. How would I prove that $$\int_{-1}^1 \frac{T_h(x)T_i(x)}{\sqrt{1-x^2}}dx=0, h \neq i$$
I can use $x=\cos(\theta)$, but I don't understand how to prove, and what this really means. Isn't it trivial because they're orthogonal?
Can anyone clarify?
Thank you. 

Using Tschebyscheff differential equation in it's autoadjoint form we have, for $T_n$ and $T_m$
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right\} + \frac{n^2}{\sqrt{1-x^2}} T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\} + \frac{m^2}{\sqrt{1-x^2}} T_m &= 0
\end{align}
Multiplying the first equation by $T_m$ and the second by $T_n$ and substracting them
\begin{align}
\frac{d}{dx}\left\{\sqrt{1-x^2} T_n'\right \}T_m - \frac{d}{dx}\left\{\sqrt{1-x^2} T_m'\right\}T_n + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0 \\
\frac{d}{dx}\left\{\sqrt{1-x^2} (T_n'T_m - T_m'T_n)\right\} + \frac{n^2 - m^2}{\sqrt{1-x^2}} T_m T_n &= 0
\end{align}
Now, the wronskian $W(T_n,T_m)$ is
$$
W(T_n,T_m) = T_n T_m' - T_n' T_m = \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{\sqrt{1-x^2}}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \int_{-1}^1 \frac{d}{d x} \left\{\frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2}\right\}dx
$$
If $n \neq m$, it's easy to see that the integral is zero. If $n \to m$, we have that
$$
\lim_{n \to m} \frac{m \cos(n \theta) \sin(m \theta) - n \sin(n \theta) \cos (m\theta)}{n^2 - m^2} = \begin{cases}-\frac{\theta}{2}, & m \neq 0, \\ \\ - \theta, & m = 0 \end{cases}
$$
and then
$$
\int_{-1}^1 \frac{T_n(x) T_m(x)}{\sqrt{1-x^2}} dx = \begin{cases} 0, & m \neq n, \\ \\
\frac{\pi}{2} & m = n \neq 0, \\ \\ \pi & m = n = 0. \end{cases}
$$
