reduction formulas

To obtain a reduction formula for ${\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}$ :

- Split off $\sin x$ to integrate by parts

$={\displaystyle \int {\sin }^{n-1}x{\cos }^{m}x\sin xdx}$

Take $u={\sin }^{n-1}x{\cos }^{m}x,\text{ and }dv=\sin xdx$ So $du=[(n-1){\sin }^{n-2}x{\cos }^{m+1}x-m{\sin }^{n}x{\cos }^{m-1}x]dx,\text{ and }v=-\cos x$
- Then simplify to get

$=-{\sin }^{n-1}x{\cos }^{m+1}x+(n-1){\displaystyle \int {\sin }^{n-2}x{\cos }^{m+2}xdx}-m{\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}$
- Now use the identity ${\sin }^{2}x+{\cos }^{2}x=1$ in the middle term and simplify to get

$=-{\sin }^{n-1}x{\cos }^{m+1}x+(n-1){\displaystyle \int {\sin }^{n-2}x{\cos }^{m}xdx}-(n-1){\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}-m{\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}$
- Take the last two integrals to the left side:

$[1+(n-1)+m]{\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}=-{\sin }^{n-1}x{\cos }^{m+1}x+(n-1){\displaystyle \int {\sin }^{n-2}x{\cos }^{m}xdx}$
- Since $[1+(n-1)+m]=m+n$ divide both sides by $m+n$ and hence

${\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}=-{\displaystyle \frac{{\sin }^{n-1}x{\cos }^{m+1}x}{m+n}}+{\displaystyle \frac{n-1}{m+n}}{\displaystyle \int {\sin }^{n-2}x{\cos }^{m}xdx}$

Using the exact same method but instead of splitting off $\sin x$ , one can split off $\cos x$ and follow similar procedure to obtain another reduction formula:

${\displaystyle \int {\sin }^{n}x{\cos }^{m}xdx}={\displaystyle \frac{{\sin }^{n+1}x{\cos }^{m-1}x}{m+n}}+{\displaystyle \frac{m-1}{m+n}}{\displaystyle \int {\sin }^{n}x{\cos }^{m-2}xdx}$