Determine an integral with legendre polynomials 
Determine the following integral $\int_{-1}^{1}x^2P_{2n-1}(x)dx$

I am unsure how to go about this. I understand that $P_{2n-1}=\frac{1}{2^{2n-1}(2n-1)!}\frac{d^{2n-1}}{dx^{2n-1}}(x^2-1)^{2n-1}$
and $\sum_{n=0}^{\infty}P_n(x)r^n=(1-2rx+r^2)^{-\frac{1}{2}}$
However I am unsure how to relate these to the integral?
 A: Hint: $2n-1$ is odd, and the interval of integration is symmetric around the origin.
A: Followin the hint of Achille Hui we have $$ \int_{-1}^{1}x^{2}P_{2n-1}\left(x\right)dx\stackrel{x\rightarrow-x}{=}-\int_{-1}^{1}x^{2}P_{2n-1}\left(x\right)dx
 $$ hence $$\int_{-1}^{1}x^{2}P_{2n-1}\left(x\right)dx=0.
 $$ Another way is integrating by parts $$I=\int_{-1}^{1}x^{2}P_{2n-1}\left(x\right)dx=\left.\frac{x^{2}\left(P_{2n}\left(x\right)-P_{2n-2}\left(x\right)\right)}{4n-1}\right|_{-1}^{1}-\frac{2}{4n-1}\left(\int_{-1}^{1}xP_{2n}\left(x\right)dx-\int_{-1}^{1}xP_{2n}\left(x\right)dx\right)
 $$ $$=-\frac{2}{4n-1}\left(\int_{-1}^{1}xP_{2n}\left(x\right)dx-\int_{-1}^{1}xP_{2n}\left(x\right)dx\right)=-\frac{2}{4n-1}\left(I_{1}-I_{2}\right)
 $$ and integrate by parts again $$I_{1}=\left.-\frac{2x\left(P_{2n+1}\left(x\right)-P_{2n-1}\left(x\right)\right)}{\left(4n-1\right)\left(4n+1\right)}\right|_{-1}^{1}+\frac{2}{\left(4n-1\right)\left(4n+1\right)}\left(\int_{-1}^{1}P_{2n+1}\left(x\right)dx-\int_{-1}^{1}P_{2n-1}\left(x\right)dx\right)=0
 $$ and the same holds for $I_{2}$ so the integral is zero.
A: A simpler approach is to notice that
$$ x^2 = \frac{2}{3} P_{2}(x) + \frac{1}{3} P_0(x) \tag{1}$$
then exploit the orthogonality of Legendre polynomials:
$$ \int_{-1}^{1}P_n(x)P_m(x)\,dx = \frac{2\delta(n,m)}{2n+1}.\tag{2}$$
Since $x^2$ is a linear combination of Legendre polynomials with even indices, $x^2$ is orthogonal to every Legendre polynomial of the form $P_{2n-1}(x)$ and the wanted integral is just $\color{red}{\large 0}$.
