Show that:
$$\int_{-1}^{1}(1-x^2)P_l'(x)P_m'(x)dx=\frac{2l(l+1)}{2l+1}\delta_{lm}.$$
So here's what I have: $$[(1-x^2)P_n'(x)]'=-n(n+1)P_n(x)$$ (from the original DiffEq), and the orthogonality relation $$\int_{-1}^{1}P_m(x)P_n(x)dx=\frac{2}{2n+1}\delta_{mn}.$$ Now, integrating by parts gives:
$$\int_{-1}^{1}(1-x^2)P_l'(x)P_m'(x)dx=\left.(1-x^2)P_l'(x)P_m(x)\right|_{-1}^{1}-\int_{-1}^{1}-l(l+1)P_l(x)P_m(x)dx$$
$$=\left.(1-x^2)P_l'(x)P_m(x)\right|_{-1}^{1}+\frac{2l(l+1)}{2l+1}\delta_{lm}.$$
I think this is the right way to go about this problem, but I don'tknow how to get rid of that $(1-x^2)P_l'(x)P_m(x)$ term. Thoughts?
In fact, term $(1-x^2)P_l'(x)P_m(x)$ should be $(1-x^2)P_l'(x)P_m(x)|_{x=-1}^{x=1}$ which naturally gives $0$.
General formula:
$$\int_a^b uv' = [uv]|_a^b - \int_a^b vu'$$