Integration of product of two Legendre polynomial

I am trying to understand the Integration of the product of two Legendre polynomials.

Given,

$$I=\int_{-1}^{1} P_{l}(x)P_{l'}(x)\mathrm{d}x~$$

Suppose $$x = \cos(\theta)$$, then our $$I$$ becomes

$$I=\int_{0}^{\pi} P_{l}(\cos(\theta))P_{l'}(\cos(\theta))\sin(\theta)\mathrm{d}\theta~$$

Since Legendre polynomials are orthogonal to each other, thus at $$l \neq l^{'}$$, $$I = 0.$$

So now we consider what happens at $$l = l'$$:

Our Integral becomes as follows:

$$I=\int_{0}^{\pi} P_{l}(\cos(\theta))P_{l}(\cos(\theta))\sin(\theta)\mathrm{d}\theta~ = \int_{0}^{\pi} (P_{l}(\cos(\theta)))^2\sin(\theta)\mathrm{d}\theta~$$

This is where I get lost. Griffith's Introduction to Electrodynamics, page no. 144, eq. 3.68, tells that for $$l=l^{'}$$, $$I = \frac{2}{2l+1}$$, but I am unable to work out that integral. Here is my attempt.

We know from Rodrigues formula that,

$$P_l(x)\equiv \frac{1}{2^l l!}\left(\frac{d}{dx}\right)^l(x^2-1)^l$$

Subsituting $$x = \cos(\theta)$$ we get,

$$P_l(\cos(\theta))\equiv \frac{1}{2^l l!}\left(\frac{d}{d\theta}\right)^l((\cos(\theta))^2-1)^l$$

Now we substitute $$P_l$$ in $$I$$, and we get:

$$I= \int_{0}^{\pi} \left(\frac{1}{2^l l!}\left(\frac{d}{d\theta}\right)^l((\cos(\theta))^2-1)^l\right)^2\sin(\theta)\mathrm{d}\theta~$$

If $$l$$ is even (an assumption I am making to make my life easy),

$$I= \left(\frac{1}{2^l l!}\right)^2\int_{0}^{\pi} \left(\left(\frac{d}{d\theta}\right)^l(\sin(\theta))^{2l}\right)^2\sin(\theta)\mathrm{d}\theta~$$

At this point, it just seems I have complicated the math even more. Can someone help me to reach the final answer? Thanks!

I think that $$x=\cos\theta$$ is an off-road way, but Rodrigues' formula is a good one:
$$\frac{d^n}{dx^n}P_n(x)=\frac1{2^n n!}\frac{d^{2n}}{dx^{2n}}(x^2-1)^n=\frac{(2n)!}{2^n n!},$$ so that integration by parts ($$n$$ times) gives $$\int_{-1}^1 P_n^2(x)\,dx\stackrel{\color{Violet}{!}}{=}\frac{(-1)^n}{2^n n!}\int_{-1}^1(x^2-1)^n\frac{d^n}{dx^n}P_n(x)\,dx=\frac{(2n)!}{2^{2n}n!^2}\int_{-1}^1(1-x^2)^n\,dx,$$ and the integral can be computed using $$x=1-2t$$ and the beta function: $$\int_{-1}^1(1-x^2)^n\,dx=2^{2n+1}\int_0^1 t^n(1-t)^n\,dt=\frac{2^{2n+1}n!^2}{(2n+1)!}.$$