Finding the value of $\int^{1}_{0}(1+x)^{m}(1-x)^{n} \,\mathrm d x$ 
Find the value of $$\displaystyle \int^{1}_{0} (1+x)^{m}(1-x)^{n} \,\mathrm d x$$ where $m, n \geq 1$ and $m, n \in \mathbb{N}$.

Let 
$$\displaystyle I_{m,n} = \int^{1}_{0}(1+x)^m(1-x)^n \,\mathrm d x = \int^{1}_{0}(2-x)^m\cdot x^n \,\mathrm d x$$
Put $x=2\sin^2 \theta$ and $dx = 4\sin \theta \cos \theta \,\mathrm d \theta$ and changing limits, We get 
$$\displaystyle I_{m,n} = 2^{m+n+2}\int^{\frac{\pi}{2}}_{0}\cos^{2m+1}\theta\cdot \sin^{2n+1}\theta \,\mathrm d \theta$$
How can I form  a recursive relation? Could someone help me? Thanks.

By parts,
$$I_{m,n}:=\int_0^1(1+x)^m(1-x)^ndx\\
=\left.\frac{(1+x)^{m+1}(1-x)^n}{m+1}\right|_0^1+\frac n{m+1}\int_0^1(1+x)^{m+1}(1-x)^{n-1}dx.$$
This gives you the recurrence relation
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}I_{m+1,n-1}.$$
From this,
$$I_{m,n}=-\frac1{m+1}+\frac n{m+1}\left(-\frac1{m+2}+\frac{n-1}{m+2}I_{m+2,n-2}\right)\\
=-\frac1{m+1}-\frac n{(m+1)(m+2)}+\frac{n(n-1)}{(m+1)(m+2)}I_{m+2,n-2}\\
=-\frac1{m+1}-\frac n{(m+1)(m+2)}-\frac{n(n-1)}{(m+1)(m+2)(m+3)}+\frac{n(n-1)(n-2)}{(m+1)(m+2)(m+3)}I_{m+3,n-3}\\
=\cdots\\
=-\sum_{k=0}^n\frac{m!n!}{(m+k+1)!(n-k)!}+\frac{n!}{(m+n+1)!}I_{m+n,0}.$$
The final integral is elementary, $\dfrac1{m+n+1}$. I don't know how to simplify the summation.
