Derivation of trig reduction formula

J

jwpaine

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

For ∫sinnxdx\displaystyle \int sin^{n}xdx we have ∫sinnxdx=−1nsinn−1xcosx+n−1n∫sinn−2xdx\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
 
Yes, JW, it is done by using parts.

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

∫sinn(x)dx=−sinn−1(x)cos(x)+(n−1)∫sinn−2(x)cos2(x)dx\displaystyle \int{sin^{n}(x)}dx=-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)cos^{2}(x)}dx

=−sinn−1(x)cos(x)+(n−1)∫sinn−2(x)(1−sin2(x))dx\displaystyle =-sin^{n-1}(x)cos(x)+(n-1)\int{sin^{n-2}(x)(1-sin^{2}(x))}dx

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

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

=∫sinn(x)dx=−sinn−1(x)cos(x)n+n−1n∫sinn−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}
 
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