Legendre Polynomials! I need some help to show that $\int_{-1}^{1}P_n^2(x)\ dx = \frac{2}{2n+1}$ where $P_n$ is the $n^{th}$ Legendre's Polynomial.
I've already done some calculations, so I would be very grateful whether someone just showed me why:
$\int_0^1 s^n(1-s)^n ds = \frac{(n!)^2}{(2n+1)!}$
It would be enough :-)
P.S. The "reduction formula" can be used... (?)
 A: To start, note that the identity you are trying to prove is related to: 
$$\mathrm{B}(a, b) = \int_0^1 s^{a-1}(1-s)^{b-1}\,ds = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
(Your particular scenario is $\mathrm{B}(n, n)$.)
I'm sure there are many proofs of this, but one that I know relates to the convolution integral and Laplace transforms.

Declarations: Let $f(t) = t^{a-1},\,g(t) = t^{b-1}$.  Also, let $h(t) = (f*g)(t)$, where $*$ is convolution.  Finally, designate the Laplace transform of $f(t),\,g(t),\,h(t)$ as $F(s),\,G(s)\,H(s)$, respectively.

Note that $H(s) = F(s)G(s)$. (This is a property of convolution and the Laplace transform.)
By deriving or looking up the Laplace transforms of $f$ and $g$, we see that:
\begin{align}
H(s) &= F(s)G(s) \\
&= \frac{\Gamma(a)}{s^a}\frac{\Gamma(b)}{s^b} \\
&= \frac{\Gamma(a)\Gamma(b)}{s^{a+b}}
\end{align}
Taking the inverse transform:
\begin{align}
h(t) &= \mathcal{L}^{-1}\left\{\frac{\Gamma(a)\Gamma(b)}{s^{a+b}}\right\}\\
&= \Gamma(a)\Gamma(b)\mathcal{L}^{-1}\left\{\frac{1}{s^{a+b}}\right\}\\
&= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left\{\frac{\Gamma(a+b)}{s^{a+b}}\right\}\\
&= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}t^{a+b-1}
\end{align}
By definition of convolution, we have that:
$$h(t) = (f*g)(t) = \int_0^t r^{a-1}(t-r)^{b-1}\,dr$$
But, we know $h(t)$ from the inverse transform.  This leads us to the identity:
$$\int_0^t r^{a-1}(t-r)^{b-1}\,dr = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}t^{a+b-1}$$

For the Beta function, we substitute $t=1$, obtaining:
  $$\mathrm{B}(a,b) = \int_0^1 r^{a-1}(1-r)^{b-1}\,dr = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

Thus, for your case:
\begin{align}
\int_0^1 r^{n}(1-r)^{n}\,dr &= \frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma((n+1)+(n+1))}\\
&= \frac{n!n!}{(2n+1)!}\\
&= \frac{(n!)^2}{(2n+1)!}
\end{align}
This is a "guided" exercise in Boyce and DiPrima's Ordinary Differential Equations and Boundary Value Problems, in the Convolution Integral section of the Laplace Transforms chapter.  (I'm mentioning this to give credit to where I first saw this.)
A: Instead of calculating $I_n=\int_0^1s^n(1-s)^nds$ I propose to calculate
$J_{n,m}=\int_0^1s^n(1-s)^mds$ and then we set $m=n$. With this said we have using integration by parts, for $m>0$:
$$
J_{n,m}=\left[\frac{s^{n+1}}{n+1}(1-s)^{m}\right]_0^1+\int_0^1s^{n+1}(1-s)^{m-1}ds
$$
or
$$
J_{n,m}=\frac{m}{n+1}J_{n+1,m-1}
$$
This shows by induction that
$$
J_{n,m}=\frac{m}{n+1}\cdot \frac{m-1}{n+2}\cdots \cdot \frac{1}{n+m}J_{n+m,0}
=\frac{m!\,n!}{(m+n+1)!}
$$
Now, set $m=n$.
