Derivation of trig reduction formula

[jwpaine]

In my book we are given a long table of trig reduction formulas, but the derivation is not supplied.

For \(\displaystyle \int sin^{n}xdx\) we have \(\displaystyle \int sin^{n}xdx = -\frac{1}{n}sin^{n-1}xcosx + \frac{n-1}{n}\int sin^{n-2}xdx\)

How is this derived? Is this derived from integration by parts, some how?

Thanks,
John

 

[galactus]

Yes, JW, it is done by using parts.

Let \(\displaystyle u=sin^{n-1}(x), \;\ dv=sin(x)dx, \;\ du=(n-1)sin^{n-2}(x)cos(x)dx, \;\ v=-cos(x)\)

\(\displaystyle \int{sin^{n}(x)}dx=-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)cos^{2}(x)}dx\)

\(\displaystyle =-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)(1-sin^{2}(x))}dx\)

\(\displaystyle =-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)}dx-(n-1)\int{sin^{n}(x)}dx\)

\(\displaystyle =n\int{sin^{n}(x)}dx=-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)}dx\)

\(\displaystyle =\boxed{\int{sin^{n}(x)}dx=\frac{-sin^{n-1}(x)cos(x)}{n}+\frac{n-1}{n}\int{sin^{n-2}(x)}dx}\)

 

[jwpaine]

Ahhhh. I should have played around with it more.

Thanks!!