# How do you use integration by parts to establish the reduction formula intsin^n(x) dx = -(1/n)sin^(n-1)(x)cos(x)+(n-1)/nintsin^(n-2)(x)dx ?

Sep 19, 2014

Let $I = \int {\sin}^{n} x \mathrm{dx}$.

By pulling out one of $\sin x$'s,

$I = \int \sin x \cdot {\sin}^{n - 1} x \mathrm{dx}$

Let $u = {\sin}^{n - 1} x$ and $\mathrm{dv} = \sin x \mathrm{dx}$
$R i g h t a r r o w \mathrm{du} = \left(n - 1\right) {\sin}^{n - 2} x \cos x$ and $v = - \cos x$

by Integration by Parts,

$= - {\sin}^{n - 1} x \cos x + \left(n - 1\right) \int {\sin}^{n - 2} x {\cos}^{2} x \mathrm{dx}$

by the trig identity ${\cos}^{2} x = 1 - {\sin}^{2} x$,

$= - {\sin}^{n - 1} x \cos x + \left(n - 1\right) \int {\sin}^{n - 2} x \left(1 - {\sin}^{2} x\right) \mathrm{dx}$

$= - {\sin}^{n - 1} x \cos x + \left(n - 1\right) \int {\sin}^{n - 2} x \mathrm{dx} - \left(n - 1\right) I$

By adding $\left(n - 1\right) I$

$R i g h t a r r o w n I = - {\sin}^{n - 1} x \cos x + \left(n - 1\right) \int {\sin}^{n - 2} x \mathrm{dx}$

By dividing by $n$,

$I = - \frac{1}{n} {\sin}^{n - 1} x \cos x + \frac{n - 1}{n} \int {\sin}^{n - 2} x \mathrm{dx}$