Use integration by parts to verify the following : Using integration by parts show that: 
$\int^{1}_{-1}P_{n}\left( x\right) P_{m}\left( x\right) dx$ = $\dfrac {2}{2n+1}, m=n$ 
and 
$0$ if $m\neq n$
Where the functions are both Legendre polynomials. 
I have tried using Rodriguez formula to try and solve it, but i cant seem to solve it. 
thank 
 A: Note that the Legendre Polynomials satisfy the ordinary differential equation
$$\frac{d}{dx}\left((1-x^2)\frac{dP_n(x)}{dx}\right)+n(n+1)P_n(x)=0 \tag 1$$
with $n\ge 0$ and $P_n(1)=1$.
To show that $\{P_n(x)\}$ are orthogonal, multiply $(1)$ by $P_m(x)$, $m\ne n$ and integrate over $[-1,1]$ to obtain
$$-n(n+1)\int_{-1}^1 P_n(x)P_m(x)\,dx=\int_{-1}^1 P_m(x)\frac{d}{dx}\left((1-x^2)\frac{dP_n(x)}{dx}\right)\,dx \tag 2$$
Integrating by parts the right-hand side of $(2)$ with $u=P_m(x)$ and $v=(1-x^2)P_n'(x)$ yields
$$-n(n+1)\int_{-1}^1 P_n(x)P_m(x)\,dx=-\int_{-1}^1 (1-x^2)\frac{dP_m(x)}{dx}\frac{dP_n(x)}{dx}\,dx= \tag 3$$
Then, integrating by parts the right-hand side of $(3)$ with $u=(1-x^2)\frac{dP_m(x)}{dx}$ and $v=P_n(x)$ yields
$$\begin{align}-n(n+1)\int_{-1}^1 P_n(x)P_m(x)\,dx&=\int_{-1}^1 P_n(x)\frac{d}{dx}\left((1-x^2)\frac{dP_m(x)}{dx}\right)\,dx\\\\&= -m(m+1)\int_{-1}^1 P_n(x)P_m(x)\,dx\tag 4\end{align}$$
Since $m\ne n$, $(4)$ implies 
$$\int_{-1}^1 P_n(x)P_m(x)\,dx=0$$
as was to be shown.
To evaluate the integral $\int_{-1}^1 P_n^2(x)\,dx$ we appeal to Rodrigue's Formula
$$P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left[(x^2-1)^n\right] \tag 5$$
Using $(5)$, we can write
$$\int_{-1}^1 P_n^2(x)\,dx=\frac{1}{4^n(n!)^2}\int_{-1}^1 \frac{d^n}{dx^n}\left[(x^2-1)^n\right]\,\frac{d^n}{dx^n}\left[(x^2-1)^n\right] \,dx \tag 6$$
Now, we integrate by parts $(6)$ $n$ times to arrive at
$$\begin{align}\int_{-1}^1 P_n^2(x)\,dx&=\frac{(-1)^n}{4^n(n!)^2}\int_{-1}^1(x^2-1)^n\frac{d^{2n}}{dx^{2n}}\left[(x^2-1)^n\right]\,dx\\\\
&=\frac{(2n)!}{4^n(n!)^2}\int_{-1}^1(1-x^2)^n\,dx\\\\
&=\frac{(2n)!}{4^n(n!)^2}2\int_0^{\pi/2}\cos^{2n+1}(\theta)\,d\theta\\\\
&=\frac{2}{2n+1}
\end{align}$$
where we tacitly used the reduction formula for the cosine function to establish the last equality.
And we are done!
