Prove that $\int_{-1}^1P_n^2(x)dx=\frac{2}{2n+1}$, where $P_n(x)$ is a Legendre polynomial. Using Rodrigues' formula and integrating by parts $n$ times, prove that $$\int_{-1}^1P_n^2(x)dx=\frac{2}{2n+1}$$
where $P_n(x)$ is a Legendre polynomial.

I tried this way
Let $$f(x)=(x^2-1)^s$$ $$\int_{-1}^1f^{(s)}(x)f^{(s)}(x)dx=-\int_{-1}^1f^{(s+1)}(x)f^{(s-1)}(x)dx$$ (the identity that I have not been able to demonstrate).
Then apply this formula repeatedly and obtain 
$$2(2s)!\int_{0}^1(1-x^2)^sdx$$
I know what to use a replacement $x=\cos\theta$ and transform this integral en 
$\int_{0}^{\pi/2} \sin^{2s+1} t \;dt$
I do not know how to continue.
 A: Recall the Rodrigues' formula
$$
P_n(x)=\frac{1}{2^nn!}\frac{d^n}{dx^n}\left(x^2-1\right)^n,
$$
then
$$
\mathcal{I}(n)=\int_{-1}^1 P_n^2(x)\ dx=\frac1{4^n(n!)^2}\int_{-1}^1\frac{d^n}{dx^n}\left(x^2-1\right)^n\frac{d^n}{dx^n}\left(x^2-1\right)^n\ dx.
$$
Using IBP yields
\begin{align}
\mathcal{I}(n)&=\frac1{4^n(n!)^2}\left[\left.\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\right|_{-1}^1-\int_{-1}^1\frac{d^{n-1}}{dx^{n-1}}\left(x^2-1\right)^n\frac{d^{n+1}}{dx^{n+1}}\left(x^2-1\right)^n\ dx\right].
\end{align}
Hence IBP $n$ times yields
\begin{align}
\mathcal{I}(n)&=\frac{(-1)^n}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\frac{d^{2n}}{dx^{2n}}\left(x^2-1\right)^n\ dx\\
&=\frac{(-1)^n(2n)!}{4^n(n!)^2}\int_{-1}^1\left(x^2-1\right)^n\ dx\\
\end{align}
Setting $x=2t-1$ yields
\begin{align}
\mathcal{I}(n)
&=\frac{(2n)!}{(n!)^2}\int_{0}^1t^n\left(1-t\right)^n\ dt\\
&=\frac{(2n)!}{(n!)^2}\cdot\text{B}(n+1,n+1)\\
&=\frac{(2n)!}{(n!)^2}\cdot\frac{(n!)^2}{(2n+1)!}\\
&=(2n)!\cdot\frac{1}{(2n+1)(2n)!}\\
&=\large\color{blue}{\frac1{2n+1}},\tag{Q.E.D.}
\end{align}
where $\text{B}(x,y)$ is a Beta function.
A: $$first\ we\ know\ :\\
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\because \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n=0}^{\infty }p_{n}(x)\ t^n\ \ \ \ \ \ \ \ \ \ \ (1)\\
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\because \frac{1}{\sqrt{1-2xt+t^2}}=\sum_{m=0}^{\infty }p_{m}(x)\ t^m\ \ \ \ \ \ \ \ \ \ \ \ (2)\\
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by\  multiplication \ (1)\ , (2)\ \ \ \ , then\ we\ have\ \\
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\therefore \frac{1}{1-2xt+t^2}=\sum_{n=0}^{\infty }\sum_{m=0}^{\infty }\ p_{n}(x)\ p_{m}(x)\ \ \ \ \ \ \ , at\ n=m\\
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\\$$
$$\therefore \frac{1}{1-2xt+t^2}=\sum_{n=0}^{\infty }\ (p_{n}(x))^2\ t^{2n}\\
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by\ integrating\ from\ -1\ to 1\ \ then\ we\ have\ \\
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\therefore \int_{-1}^{1}\sum_{n=0}^{\infty }(p_{n}(x))^2dx=\frac{-1}{2t}\int_{-1}^{1}\frac{-2tdt}{1-2xt+t^2}dx=\frac{-1}{2t}ln(1-2xt+t^2)_{-1}^{1}\\
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=\frac{-1}{2t}\left [ ln(1-2t+t^2)-ln(1+2t+t^2) \right ]=\frac{-1}{2t}\left [ ln(1-t)^2-ln(1+t)^2 \right ]\\
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=\frac{-1}{t}\left [ ln(1-t) -ln(1+t)\right ]\\
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\\$$
$$=\frac{-1}{t}\left [ \frac{-t}{1} -\frac{t^2}{2}-\frac{t^3}{3}-\frac{t^4}{4}-......+(\frac{t}{1}-\frac{t^2}{2}+\frac{t^3}{3}-......\right ]\\
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=\frac{-1}{t}\left [ \frac{-2t}{1} -\frac{2t^3}{3}-.....\right ]=2\left [ 1+\frac{t^2}{3} +\frac{t^4}{5}+......\right ]\\
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=2\sum_{n=0}^{\infty }\frac{t^{2n}}{2n+1}=\sum_{n=0}^{\infty }\int_{-1}^{1}(p_{n}(x))^2\ t^{2n}\\
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by\ coffecients\ we\ have\ \ \\
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\therefore \frac{2}{2n+1}=\int_{-1}^{1}\ p_{n}(x)\ p_{m}(x)\ dx \ \ \ \ \ \ \ , n=m\\
\\$$
