Show that $T_{0}(x)=1$ ,$T_{1}(x)=x$, and $T_{2}(x)=2x^{2}-1$ are mutually orthogonal on the interval $(-1,1)$ whit respect to the weighting function $r(x)=\left (1-x^{2} \right )^{-1/2}$. These polynomials are the first three of the set of Chebyshev polynomials
Hint: Let $x=cos (\theta) $
I have trouble advancing with this problem, though it seemed quite straight forward at glance.
I've done this, but I don't know if it's okay
$\left \langle T_{0},T_{1} \right \rangle=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}(1)(x)dx$
$\left \langle T_{0},T_{2} \right \rangle=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}(1)(2x^{2}-1)dx$
$\left \langle T_{1},T_{2} \right \rangle=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}(x)(2x^{2}-1)dx$
If this problem is solved in this way, I also use the "Hint" $x=cos (\theta)$ in the squares of this problem.
$$\begin{aligned} \left \langle T_{0},T_{1} \right \rangle &=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}x \,dx=\int_0^\pi \cos \theta \,d\theta = \left.\sin \theta\right|_{-\pi}^\pi=0.\\ \left \langle T_{0},T_{2} \right \rangle&=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}(2x^2-1) \,dx=\int_0^\pi \cos 2 \theta \,d\theta = \left.\frac{\sin 2 \theta}{2} \right|_{-\pi}^\pi=0.\\ \left \langle T_{1},T_{2} \right \rangle&=\int_{-1}^{1}\left (1-x^{2} \right )^{-1/2}(2x^2-1)\, x\,dx=\int_0^\pi \cos \theta \cos 2 \theta \,d\theta = \left[\frac{\sin \theta}{2} - \frac{\sin 3\theta}{6} \right]_{-\pi}^\pi=0.\end{aligned}$$