TITLE: Prove reduction formula for $\int \cos^n (x)\sin^m (x) \, dx$
QUESTION [8 upvotes]: $$\displaystyle\int \:\sin^n\left(x\right)\cos^m\left(x\right)\mathrm  dx=\frac{\sin^{n+1}x\cos^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \:\sin^nx\cos^{m-2}x\,\mathrm dx$$
I have been trying to solve for over a week now can someone please help me.

REPLY [2 votes]: $$I_{m,n}=\int \sin^nx\cos^mx dx=\int \sin^nx\cos^{m-1}x \cos xdx$$
$$I_{m,n}=\int \sin^nx\cos^mx dx=\int \cos^{m-1}x \sin^nx\cos xdx$$
$$=\cos^{m-1}x\frac{sin^{n+1}x}{n+1}-\frac{m-1}{n+1}\int \sin^{n+2}x\cos^{m-2}x dx$$ ($\because$ Integration by parts)
$$=\cos^{m-1}x\frac{sin^{n+1}x}{n+1}-\frac{m-1}{n+1}\int \sin^{n}x\cos^{m-2}x(1-\cos^2x) dx $$
Rearrangeing the terms. We get,
 $$\displaystyle\int \:\sin^n\left(x\right)\cos^m\left(x\right)\mathrm  dx=\frac{\sin^{n+1}x\cos^{m-1}x}{m+n}+\frac{m-1}{m+n}\int \:\sin^nx\cos^{m-2}x\,\mathrm dx$$