Consider the $L^2([-1,1])$ function space with the follwoing inner product:
$$\langle u(x),v(x)\rangle =\int_{-1}^{1} u(x)\bar{v}(x) dx$$
Rodrigues's Polynomials in $[-1,1]$ are defined as:
$$R_n(x)=\frac{d^n}{dx^n} \left(x^2-1\right)^n $$
I pretend to determinate $||R_n(x)||^2=\langle R_n(x),R_n(x)\rangle$.
There's a tip for this exercise:
$$\int_{0}^{\frac{\pi}{2}} \cos(\theta)^{2n+1} d\theta=\frac{\left(2^nn!\right)^2}{(2n+1)!}$$
So, I started to solve $\frac{d^n}{dx^n} \left(x^2-1\right)^n$ and I realized that this function has the following form:
$$R_n(x)=\sum_{k=1}^n c_k(x^2-1)^k$$
I noticed that finding an expression for the coefficients $c_k$ was too much difficult, so maybe using this form couldn't help.
I tried to understand where I should use my tip. I think it could be useful when making a change of variable $x\to \cos(\theta)$, just because $-1<\cos(\theta)<1$ for $\theta\in[0,\pi]$ and the integral limits for the inner product are $-1$ and $1$.
But I couldn't solve this problem.
Using repeated integration by parts we have:
$$ \int_{-1}^{1}R_n(x)^2\,dx = (-1)^n \int_{-1}^{1}(x^2-1)^n\cdot \frac{d^{2n}}{dx^{2n}}(x^2-1)^n\,dx = (2n)!\int_{-1}^{1}(1-x^2)^n\,dx$$ and by setting $x=\sin\theta$ in the last integral we just have to apply the given tip.