Integration of $(x(1-x))^n$ I'm wondering how to find out the definite integration of 
$$\int^y_0(x(1-x))^ndx.$$
I've tried several integration techniques but couldn't make it. Would there be a closed-form representation of this besides the one related to an incomplete beta function? 
 A: Let $x=\sin^2 t$ ,so
$$I=\int 2 \sin^{2n+1} t \cos^{2n+1} t dt= \int 2^{-2n} \sin^{2n+1} 2t~ dt =2^{-(2n+1)} \int \sin^{2n+1} u ~du= J_{2n+1}$$
Next one may use the reduction formula usinf integration by parts: $\sin^n x \sin x$
$$J_{2n+1}=2^{-(2n+1)} \int \sin^{2n} u \sin u du=-2^{-(2n+1)} \sin^{2n} u \cos u + 2n~ 2^{-(2n+1)}\int \sin^{2n-1} u\cos^2 u du$$ $$ J_{2n+1}=-2^{-(2n+1)}\sin^{2n} u \cos u+n 2^{-(2n+1)}\int \sin^{2n-1} (1-\sin^2 u) du$$ $$\implies J_{2n+1}=-2^{-(2n+1)}[ -\sin^{2n} u \cos u]+\frac{n}{2} J_{2n-1}-2nJ_{2n+1},$$
$$J_{2n+1}=-\frac{2^{-(2n+1)}}{2n+1} \sin^{2n} u \cos u+ \frac{n}{2(2n+1)} J_{2n-1},J_1=-\frac{1}{2} \cos u$$
A: Assuming $x >0$ and $n >0$, you also could use
$$\int (x(1-x) )^n\,dx=\frac{x^{n+1} }{n+1}\, _2F_1(-n,n+1;n+2;x)$$ where appears the gaussian hypergeomtric function. This makes
$$\int_0^y (x(1-x) )^n\,dx=\frac{y^{n+1} }{n+1}\, _2F_1(-n,n+1;n+2;y)$$
A: $$\dfrac{d(x^a(1-x)^b)}{dx}=ax^{a-1}(1-x)^b-bx^a(1-x)^{b-1}$$
Integrate both sides with respect to $x,$
$$x^a(1-x)^b+K=aI(a-1,b)-bI(a,b-1)$$ where $$I(m,n)=\int x^m(1-x)^ndx$$
Observe that for positive integer $a,b$ we are actually reducing the exponent of $1-x$ and increasing that if $x$ until the earlier reaches $0$
Start with $a-1=b=n$
A: There is no compact closed-form expression for the incomplete Beta function so the answer is no.

For integer $n$, you can obviously use the binomial development,
$$\int_0^y x^n\sum_{k=0}^n(-1)^k\binom nk x^k\,dx=\sum_{k=0}^n(-1)^k\binom nk \frac{y^{k+n+1}}{k+n+1}$$
which is a closed-form expression, but probably not wat you hope. For fractional $n$, the generalized binomial development should also work.
