To prove that Legendre polynomials are orthogonal, i.e.
$$
\int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \frac{2\,\delta_{m,n}}{2n+1}
$$
you can use Rodrigues' formula (eq 3.62 in Griffiths):
$$
P_n(x) = \frac{1}{2^n n!}\left(\frac{\text{d}}{\text{d}x}\right)^n (x^2-1)^n.
$$
We get
$$
I_{mn} = \int_{-1}^1 P_n(x)P_m(x)\,\text{d}x =
\int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^n \!X^n\,\text{d}x,
$$
where I used the short-hand notation $X = x^2-1$. Without loss of generality, we can assume that $n\geqslant m$. Let us integrate this by parts:
$$
\begin{multline}
I_{mn} = \left[\frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^m \!X^m \,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\right]_{-1}^1 \\-
\int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x.
\end{multline}
$$
However, since the derivative
$$
\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} (x^2-1)^n
$$
will contain a factor $(x^2-1)$, the integrated part will vanish, so that only the integral remains:
$$
I_{mn} = -\int_{-1}^1 \frac{1}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+1} \!X^m\,\left(\frac{\text{d}}{\text{d}x}\right)^{n-1} \!X^n\,\text{d}x.
$$
In this manner, we can integrate by parts $n$ times, until we obtain
$$
I_{mn} = (-1)^n\int_{-1}^1 \frac{(x^2-1)^n}{2^{m+n} m!\,n!}\left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m\,\text{d}x.
$$
Now,
$$
(x^2-1)^m = x^{2m} - mx^{2m-2} + \ldots
$$
So that, if $n>m$,
$$
\left(\frac{\text{d}}{\text{d}x}\right)^{m+n} \!(x^2-1)^m = 0,
$$
and $I_{mn}=0$.
On the other hand, if $n=m$, we get
$$
\left(\frac{\text{d}}{\text{d}x}\right)^{n+n} \!(x^2-1)^n = (2n)!
$$
and
$$
I_{nn} = (-1)^n\int_{-1}^1 \frac{(2n)!}{2^{2n} n!\,n!}(x^2-1)^n\,\text{d}x.
$$
Substituting $x$ with $y = (x+1)/2$, we can rewrite this as
$$
I_{nn} = 2\int_0^1 \frac{(2n)!}{n!\,n!}y^n(1-y)^n\,\text{d}y,
$$
which is a beta function:
$$
I_{nn} = 2\frac{(2n)!}{n!\,n!}B(n+1,n+1) = 2\frac{(2n)!}{n!\,n!}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)} = 2\frac{(2n)!}{(2n+1)!} = \frac{2}{2n+1},
$$
and this concludes the proof.
With $x=\cos\theta$, you can write it as
$$
\begin{multline}
\int_{-1}^1 P_n(x)P_m(x)\,\text{d}x = \int_{-1}^1 P_n(\cos\theta)P_m(\cos\theta)\,\text{d}\cos\theta \\= \int_{0}^\pi P_n(\cos\theta)P_m(\cos\theta)\sin\theta\,\text{d}\theta.
\end{multline}
$$
Because of the orthogonality property, you can find the coefficients by integrating $V(r,\theta)$ with each Legendre polynomial:
$$
\int_{0}^\pi V(r,\theta)P_l(\cos\theta)\sin\theta\,\text{d}\theta = \frac{2}{2l+1}\left(A_lr^l + \frac{B_l}{r^{l+1}}\right).
$$
