In Wikipedia, it is mentioned that for Chebyshev polynomials "An important and convenient property of the Tn(x) is that they are orthogonal with respect to the inner product":
$\langle f(x), g(x)\rangle=\int_{-1}^{1} f(x) g(x) \frac{\mathrm{d} x}{\sqrt{1-x^{2}}}$
I don't understand the statement. Usually a function is orthogonal with respect to another function, however, what does it mean when a function is orthogonal to it's inner product?
Think first of orthogonality of standard discrete vectors in Euclidean space, ${\bf a}$ and ${\bf b}$. The orthogonality condition is simply ${\bf a}^t {\bf b} = a_1 b_1 + a_2 b_2 + \cdots + a_n b_n = 0$. Now imagine that we have functions, $f(x)$ and $g(x)$. The natural extension of the orthogonality condition would be:
$$f(x_1) g(x_1) + f(x_2) g(x_2) + \cdots f(x_n) g(x_n) = 0.$$
Now let the number of $x$s go to infinity, i.e., $n \to \infty$, and the difference between sampled steps approaches zero. In the limit, you have the integral definition stated in your question, $\int f(x) g(x) dx = 0$.
(The other factor is just for normalization, based on the definition of your class of functions.)