# reduction formulas

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

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

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

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

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

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

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

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

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:

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

Title reduction formulas ReductionFormulas 2013-03-22 17:37:06 2013-03-22 17:37:06 curious (18562) curious (18562) 9 curious (18562) Definition msc 26A36 powers of sines and cosines integration of trigonometric functions TrigonometricFormulasFromDeMoivreIdentity