Determine the following integral
$$\int_{-1}^{1}x^{2}P_{2n-1}\left(x\right) dx$$
Using the generating function and the fact that $\left(1-2rx+r^2\right)^{-1/2}=\sum^{\infty}_{n=0}P_{n}\left(x\right)r^{n}$ in previous questions I have calculated, $P_{n}\left(1\right)=1$ and $P_{n}\left(-1\right)=\left(-1\right)^{n}$. However I'm not sure how to calculate $P_{2n-1}\left(x\right)$. Would it be by using recurrence relations?
Legendre polynomials have the parity of their index. This means that polynomials with even index are even functions, and with odd index (as in your case) are odd functions. An odd function times an even function ($x^2$) produces an odd function, which (integrated over a symmetric interval, like $[-1,1]$) gives $0$. So your integral is zero for all $n$, without the need of any calculation.