$ \int_{0}^{1}x^m.(1-x)^{15-m}dx$ where $m\in \mathbb{N}$ $=\displaystyle \int_{0}^{1}x^m.(1-x)^{15-m}dx$ where $m\in \mathbb{N}$
My Try:: Put $x=\sin^2 \theta$ and $dx = 2\sin \theta.\cos \theta.d\theta$ and changing limit, We Get
$ = \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2m}\theta.\cos^{30-2m}\theta.2\sin \theta.\cos \theta d\theta$
$ = \displaystyle 2\int_{0}^{\frac{\pi}{2}} \sin^{2m+1}\theta.\cos^{31-2m}\theta d\theta$
Now How can i proceed after that.
Is there is any method to Calculate the Given Integral Then plz explain here.
Thanks

$$f(m,n)=\int_0^1 x^m(1-x)^ndx$$
Repeated partial integration on the right hand side reveals that:
$$(m+1)f(m,n)=n\,f(m+1,n-1)$$
$$(m+2)(m+1)f(m,n)=n(n-1)\,f(m+2,n-2)$$
$$\cdots$$
$$(m+n)\cdots (m+1)\,f(m,n)=n!\,f(m+n,0)$$
$$\text{i.e}$$
$$(m+n)!f(m,n)=n!\,m!\,f(m+n,0)$$
Since
$$f(m+n,0)=\int_0^1 t^{m+n}dt=\frac{1}{m+n+1}$$
We have:
$$f(m,n)=\frac{n!\,m!}{(m+n+1)!}$$
Now simply let $n=15-m$
