# Reduction formula

$I_{m,n}=\int_0^{\pi/2} \sin^m \theta \cos^n \theta d \theta$. Prove that $I_{m,n}=I_{n,m}$, and that $I_{m,n}=\frac{m-1}{m+n}I_{m-2,n}$, for $m>1$.

I understand there is a property with sin and cos that proves the first part, and I have to take out sin^2 for the second part.

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I think this was asked before. math.stackexchange.com/q/323984/8581 –  Babak S. Mar 10 '13 at 17:05
@Babak: The $I_{m,n}=I_{n,m}$ part isn't in that other question. –  joriki Mar 10 '13 at 17:12

$$I_{m,n}=\int\limits_0^{\pi/2}\sin^m t\cos^n t\,dt=\int\limits_0^{\pi/2}\cos^m\left(\frac{\pi}{2}-t\right)\sin^n\left(\frac{\pi}{2}-t\right)t\,dt=$$
(and making the substitution $\,u:=\pi/2-t\Longrightarrow -du=dt$ ,we get)
$$=\int\limits_{\pi/2}^0\cos^mu\sin^nu(-du)=\int\limits_0^{\pi/2}\sin^nu\cos^mu\,du=I_{n,m}$$