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

Sep 11, 2014

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

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

By Integration by Parts,
$I = {\cos}^{n - 1} x \sin x + \left(n - 1\right) \int {\cos}^{n - 2} x {\sin}^{2} x \mathrm{dx}$
By ${\sin}^{2} x = 1 - {\cos}^{2} x$,
$I = {\cos}^{n - 1} x \sin x + \left(n - 1\right) \int {\cos}^{n - 2} x \left(1 - {\cos}^{2} x\right) \mathrm{dx}$
By splitting the last integral,
$I = {\cos}^{n - 1} x \sin x + \left(n - 1\right) \int {\cos}^{n - 2} x \mathrm{dx} - \left(n - 1\right) I$
By adding $\left(n - 1\right) I$,
$n I = {\cos}^{n - 1} x \sin x + \left(n - 1\right) \int {\cos}^{n - 2} x \mathrm{dx}$
By dividing by $n$,
$I = \frac{1}{n} {\cos}^{n - 1} x \sin x + \frac{n - 1}{n} \int {\cos}^{n - 2} x \mathrm{dx}$