Prove by induction of integration functions If $J_n= \int_{0}^{1}t^n(1-t)^ndt$, then $2(1+2n)J_n = nJ_{n-1}$ and $J_0 = 1$.
[Hint: integration by parts]
Here is what I have done so far. Am I correct so far? How can I continue to prove it? 
$J_n= \int_{0}^{1}t^n(1-t)^ndt$
    = $[(1-t)^n\cdot \frac{t^{n+1}}{n+1}]\big|_0^1$ $- \int_{0}^{1}\frac{t^{n-1}}{n+1}\cdot n(1-t)^{n-1}(-1)dt $
$= \frac{n}{n+1}\int_{0}^{1}t^{n+1}(1-t)^{n-1}dt 
 = \frac{n}{n+1}\int_{0}^{1}t^n(1-(1-t))\cdot(1-t)^{n-1}dt$
$= \frac{n}{n+1}\left(\int_{0}^{1}t^n(1-t)^{n-1}dt - \int_{0}^{1}t^n(1-t)^ndt\right) $
$ =\frac{n}{n+1}\left(\int_{0}^{1}t^{n-1}(1-(1-t))(1-t)^{n-1}dt - J_n\right)$
 A: What you have done is correct. If you multiply both sides by $n+1$ and transfer the last term to the left side  you will almost get what you want except that you need the following: $\int_0^{1}t^{n-1}(1-t)^{n-1}(2t-1)dt=0$. This equation follows from the fact that the integral of an odd function from $-\frac 1  2 $ t0 $+\frac 1  2$ is $0$; to use this fact just make the substitution $s=t-\frac 1  2$ in  $\int_0^{1}t^{n-1}(1-t)^{n-1}(2t-1)dt$.
A: $$J_n = \int_0^1t^n(1-t)^ndt$$
Using Integration by parts,
$$J_n = \frac{t^{n+1}(1-t)^n}{n+1}\vert_0^1+\frac{n}{n+1}\int_0^1t^{n+1}(1-t)^{n-1}dt$$
$$J_n = \frac{n}{n+1}\int_0^1t^{n}(1-t)^{n-1}(t)dt$$
$$J_n = \frac{n}{n+1}\int_0^1t^{n}(1-t)^{n-1}(t-1+1)dt$$
$$J_n = -\frac{n}{n+1}\int_0^1t^{n}(1-t)^{n}dt + \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt = -\frac{n}{n+1}J_n + \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt$$
$$\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^ndt \cdots(1)$$
Using the property that $\int_0^af(x)dx=\int_0^af(a-x)dx$ for $a=1$, we get
$$\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^nt^{n-1}dt \cdots(2)$$
Adding equations $(1)$ and $(2)$, 
$$2\frac{2n+1}{n+1}J_n = \frac{n}{n+1}\int_0^1(1-t)^{n-1}t^{n-1}dt = \frac{n}{n+1}J_{n-1}$$
$$\therefore 2(2n+1)J_n = nJ_{n-1}$$
which proves the desired result. Hope this helps.
