Integrating Chebyshev polynomial of the first kind I'm trying to evaluate the integral of the Chebyshev polynomials of the first kind on the interval $-1 \leq x \leq 1 $ .
My idea is to use the closed form 
$$T_n(x) = \frac{z_1^n + z_2^{-n} }{2}$$ 
where $z_1 = (x + \sqrt{x^2 - 1})$ and $z_2 = (x - \sqrt{x^2 - 1})$,
giving the following integral:
$$
\int _{-1}^{1}\!1/2\, \left( x+\sqrt {-1+{x}^{2}} \right) ^{n}+1/2\,
 \left( x-\sqrt {-1+{x}^{2}} \right) ^{n}{}{dx}
$$
I'm stuck at integrating $z_1$ and $z_2$. I tried integrating by parts $n$ times, but i'm looking for a general formula. My calculus is pretty rusty so i'm not sure if this is the way to go. Any tips? 
Thanks a lot.
 A: 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:
$$\begin{align}\int T_n(x) dx = & \int T_n(\cos\theta) d\cos \theta\\
= & -\int \cos(n\theta)\sin\theta d\theta\\
= & -\frac12 \int \left(\sin((n+1)\theta) - \sin((n-1)\theta)\right)d\theta\\
= & \frac12 \left(\frac{\cos((n+1)\theta)}{n+1} - \frac{\cos((n-1)\theta)}{n-1}\right) + \text{const.}\\
= & \frac12 \left(\frac{T_{n+1}(x)}{n+1} - \frac{T_{n-1}(x)}{n-1}\right) + \text{const.}
\end{align}
$$
A: Wikipedia has a nice article on the Chebychev polynomials: http://en.wikipedia.org/wiki/Chebyshev_polynomials.
In particular, there is this:
$\int T_n(x) dx = \frac{n T_{n+1}(x)}{n^2-1}-\frac{x T_n(x)}{n-1}$.
A: An alternative easy way to do this would be to convert everything to the complex exponential. Given
$$T_n(\cos\theta) = \cos(n\theta) = \frac{1}{2}\left( e^{in\theta} + e^{-in\theta} \right) $$
and since $\frac{d\cos\theta}{d\theta} = \frac{i}{2}\left( e^{i\theta} - e^{-i\theta} \right)$ we have
$$\begin{align}\int T_n(x)dx & = \int T_n(\cos\theta)d\cos\theta \\
& = \int \cos(n\theta)d\cos\theta \\
&= \frac{i}{4}\int \left( e^{in\theta} + e^{-in\theta} \right)\left( e^{i\theta} - e^{-i\theta} \right)d\theta \end{align}$$ 
which is straightforward to integrate. At the end of the day you will get 
$$\int T_n(x)dx  = \frac{1}{2}\left(\frac{T_{n+1}}{(n+1)}- \frac{T_{n-1}}{(n-1)} \right) $$ 
after converting the complex exponential back to the trigonometric form and using the definition of the $n$-th Chebyshev polynomial as given above.
