<s>[INST] Trouble finding the solution of $\int_0^{\pi/2}\sin^mx\cdot\cos^nx\,dx$ from the recurrence relation 
Prove
  \begin{align}
& \int_0^{\pi/2}\sin^mx \cos^nx\,dx \\[12pt]
= {} & \begin{cases}\dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots}\cdot\dfrac{\pi}{2} & m,n\text{ even integers}\\[10pt]
\dfrac{[(m-1)(m-3)\cdots][(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots} & \text{elsewhere}
\end{cases}
\end{align}

My Attempt
$$
I_{m,n}=\int_0^{\pi/2}\sin^mx\cdot\cos^n x\,dx=\int_0^{\pi/2}\sin^{m-1}x \cos^nx \sin x \, dx\\
\int\cos^nx\cdot \sin x\,dx=\dfrac{-\cos^{n+1}x}{n+1}+K\\
I_{m,n}=\bigg[\sin^{m-1}x\cdot \dfrac{-\cos^{n+1}x}{n+1}\bigg]_0^{\pi/2}-(m-1)\int_0^{\pi/2}\sin^{m-2}x\cdot \cos x\cdot \dfrac{-\cos^{n+1}x}{n+1}\,dx\\
=0+\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m-2}x(1-\sin^2x)\cdot dx\\
=\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m-2}x\,dx-\frac{m-1}{n+1}\int_0^{\pi/2}\cos^nx\cdot\sin^{m}x\,dx\\
I_{m,n}=\frac{m-1}{n+1}\cdot I_{m-2,n}-\frac{m-1}{n+1}\cdot I_{m,n}\implies
\color{blue}{\boxed{I_{m,n}=\frac{m-1}{m+n}\cdot I_{m-2,n}}}
$$
case 1: $m,n$ even:
$$
I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots 2}\cdot I_{0,n}\\
I_{0,n}=\int_0^{\pi/2}\cos^nx.dx=\frac{(n-1)(n-3)...1}{n(n-2)...2}.\frac{\pi}{2}\quad;\text{ since $n$ is even}\\
I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots 2} \cdot \frac{(n-1)(n-3)...2}{n(n-2)...2}.\frac{\pi}{2}\\
=\frac{[(m-1)(m-3)\cdots 1][(n-1)(n-3)...2]}{[(m+n)(m+n-2)\cdots 2]\color{red}{[n(n-2)...2]}}.\frac{\pi}{2}
$$ 
Why am I not getting the solution given in my reference, more precisely, why do I get the additional term, marked in red, in the final solution ?
How do I reach the final solution from the recurrence relation ?
Similar question is asked before in Reduction formula for integral $\sin^m x \cos^n x$ with limits $0$ to $\pi/2$, but it does not seem to get the same recurrence relation as in my attempt.
Note: I am having trouble proceeding from the recurrence relation, post mentioned below are attempts to derive the recurrence relation, which does not address the problem, that i have already derived it in my attempt.
 [/INST] Your error is here:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{2}}\cdot I_{0,n}.$$
The correct expression is:
$$I_{m,n}=\frac{(m-1)(m-3)\cdots 1}{(m+n)(m+n-2)\cdots\color {red}{(n+2)}}\cdot I_{0,n}.$$
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