<s>[INST] 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.
 [/INST] We first express the integrand as a product of $\sin \theta$ and $\cos \theta  $ by
letting $x=\cos 2\theta $, yields
$$I = \int_{\frac{\pi}{4}}^{0}\left(2 \cos ^{2} \theta\right)^{m}\left(2 \sin ^{2} \theta\right)^{n}\left(-2 \sin ^{2} \theta\right) d \theta 
=2^{m+n+2} \int_{0}^{\frac{\pi}{4}} \cos ^{2 m+1} \theta \sin^{2n+1} \theta d\theta 
$$
Putting $s=\sin \theta$ gives $$
I=2^{m+n+2}  \int_{0}^{\frac{1}{\sqrt{2}}}\left(1-s^{2}\right)^{m} s^{2 n+1} d s .
$$
Using Binomial Expansion gives $$
\begin{aligned}
I&=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}\left(\begin{array}{c}
m \\
k
\end{array}\right)(-1)^{k} s^{2 k} s^{2 n+1} d s\\
&=2^{m+n+2} \int_{0}^{\frac{1}{\sqrt{2}}} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) s^{2 k+2 n+1} d s\\
&=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \int_{0}^{\frac{1}{\sqrt{2}}} s^{2 k+2 n+1} d s\\
&=2^{m+n+2} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{l}
m \\
k
\end{array}\right)\left[\frac{s^{2 k+2 n+2}}{2 k+2 n+2}\right]_{0}^{\frac{1}{\sqrt{2}}}\\
&=2^{m+n+1} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \frac{1}{(k+n+1) 2^{k+n+1}}\\
&=2^{m} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}
m \\
k
\end{array}\right) \frac{1}{(k+n+1) 2^{k}}
\end{aligned}
$$
</s>