Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$ I found in this link the reduction formula
$$
\int_0^{\pi/2} \sin^m x \cos^n x \ dx =
\begin{cases}
\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2} (\frac{\pi}{2}) & m, n \text{ even}\\[5pt]
\frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)} & \text{otherwise}
\end{cases}
$$
How can I prove it?   
 A: Let $I_{m,n}=\int_0^{\pi/2} \sin^m x \cos^n x \ dx$, integrating by parts we find that 
$$
I_{m,n}=\frac{n-1}{m+1} I_{m+2,n-2} \qquad (1) \\
I_{m,n}=\frac{m-1}{n+1} I_{m-2,n+2} \qquad (2)
$$
Using (1) when $n$ is odd, 
$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 2}{(m+n-2) \cdots (m+1)}I_{m+n-1,1}=\frac{(n-1)(n-3) \cdots 2}{(m+n)(m+n-2) \cdots (m+1)} \quad (3)
$$
Interchaging $m$ and $n$ in (3) we find $I_{m,n}$ when $m$ is odd.
Then
$$
I_{m,n} = \frac{[(m-1)(m-3) \cdots (2\text{ or }1)][(n-1)(n-3) \cdots (2\text{ or }1)]}{(m+n)(m+n-2) \cdots (2\text{ or }1)}
$$
when either $m$ or $n$ are odd.
If $n$ and $m$ are even
$$
I_{m,n}=\frac{(n-1)(n-3) \cdots 1}{(m+n-1) \cdots (m+1)}I_{m+n,0}
$$
On the other hand, let $J_k=\int_0^{\pi/2} \sin^k x \ dx$. Again,integrating by parts we find:
$$
J_k=\frac{k-1}{k}J_{k-2}
$$
In particular when $k$ is even
$$
J_k=\frac{(k-1)(k-3) \cdots 1}{k(k-2) \cdots 2}\left(\frac{\pi}{2}\right)
$$
Then
$$
\begin{align}
I_{m,n}&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n-1)(m+n-3) \cdots 1} \left[\frac{(m+n-1)(m+n-3) \cdots 1}{(m+n)(m+n-2) \cdots 2} \left(\frac{\pi}{2}\right)\right]\\
&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2}\left(\frac{\pi}{2}\right)
\end{align}
$$
A: Hint- 
break the expression into $\sin^{(m-1)} x \; \{ \cos ^n x  \sin x\}$.
Let 2nd function which is in curly bracket and 1st function which is out side. Now integrate using integration by part.
