Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$

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

$$ \begin{align} I_{m,n}&=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n-1)(m+n-3) \cdots 1} \left[\frac{(m+n-1)(m+n-3) \cdots 1}{(m+n)(m+n-2) \cdots 2} \left(\frac{\pi}{2}\right)\right]\\ &=\frac{[(m-1)(m-3) \cdots 1][(n-1)(n-3) \cdots 1]}{(m+n)(m+n-2) \cdots 2}\left(\frac{\pi}{2}\right) \end{align} $$

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.