Let $P_n, P_m$ be Legendre polynomials with $n \neq m$ then: $$\int_{-1}^1 P'_m(x) P'_n (x) (1-x^2) dx=0$$
I already know that $\int_{-1}^1 P_m (x) P_n (x) dx=0$, using that $P_n$ and $P_m$ are solutions of Legendre equation then:
$$ [(1-x^2)P'_n]' +n (n+1)P_n=0 $$ $$ [(1-x^2)P'_m]' +m (m+1)P_m=0 $$
Then if we integrate $(Equation 1)P_m - (Equation 2)P_n$ we obtain:
$$\int_{-1}^{1} P_m[(1-x^2)P'_n]' - P_n [(1-x^2)P'_m]' dx + [n (n+1) - m (m+1)]\int_{-1}^1 P_n P_m dx =0 $$
If we use integration by parts we obtain that the first integral is $0$, then $\int_{-1}^1 P_m(x) P_n (x) dx=0$
I want to use a similar method to prove that $\int_{-1}^1 P'_m(x) P'_n (x) (1-x^2) dx=0$ but I don't know where to start.
We want to prove that $$\int_{-1}^1 P'_m(x) P'_n (x) (1-x^2) dx=0$$
Using integration by parts
$$u= (1-x^2)P'_m(x) \\ du = [(1-x^2)P'_m(x)]'dx$$ and $$dv=P'_n dx \\ v = P_n(x)$$
We have that $[(1-x^2)P'_m]'+m(m+1)P_m=0$ then $du = -m(m+1)P_mdx$.
$$\int_{-1}^1 P'_m(x) P'_n (x) (1-x^2) dx = (1-x^2)P'_m(x)P_n(x)\vert_{-1}^1 + m(m+1)\int_{-1}^1 P_n(x)P_m dx =0 $$