I am trying to prove the following identity:
$$\int_0^\pi P_l(\cos\theta)P_{l'}(\cos\theta)\sin\theta d\theta = \biggl\lbrace \begin{matrix}0, \ if \ l' \neq l \\ \frac2{(2l+1)}, \ if \ l'=l \end{matrix} \tag{1}$$
Where $P_l$ and $P_l'$ are defined by Rodrigues' formula
$$P_l(x) = \frac1{2^l l!} \ \partial_x^l(x^2 - 1)^l \tag{2}$$ $$P_{l'}(x) = \frac1{2^{l'} {l'}!} \ \partial_x^{l'}(x^2 - 1)^{l'} \tag{3}$$
I substituted $(2)$ and $(3)$ into $(1)$ in order to evaluate the integral for $l \neq l'$, but I am having difficulty tackling the $\partial_{\cos \theta}^l (\cos^2\theta-1)^l$ portion for both $l$ and $l'$
I tried to do the first 5 differentiations from $\partial^l$ to $\partial^{l-5}$ to find some sort of pattern but am unable to find anything useful.
Any hints or reference would be great!
First, no need to keep complicated trigonometric functions. Instead, change the variable in the integral $\cos \theta = x$.
$$\int_0^\pi P_l(\cos\theta)P_{l'}(\cos\theta)\sin\theta d\theta =\int_{-1}^1P_l(x)P_{l^\prime}(x)dx$$
Now you can use Rodrigues' formula in the integral to show the orthogonality. For instance, see this.