# Integrate cos^4x?

Apr 17, 2018

See below

#### Explanation:

As bonus: lets calculate $\int {\cos}^{n} x \mathrm{dx}$ and then we will have an special case for $n = 4$

${I}_{n} = \int {\cos}^{n} x \mathrm{dx}$ lets do it by parts

$u = {\cos}^{n - 1} x$ ; $\mathrm{dv} = \cos x \mathrm{dx}$

$\mathrm{du} = \left(n - 1\right) {\cos}^{n - 2} x \left(- \sin x\right) \mathrm{dx}$ and $v = \sin x$

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

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

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

Then we have

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

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

And we get the reduction formula

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

$n = 0 \ddots$; ${I}_{0} = \int {\cos}^{0} x \mathrm{dx} = \int \mathrm{dx} = x$
$n = 1 \ddots$ ${I}_{1} = \int \cos x \mathrm{dx} = \sin x$
$n = 2 \ddots$; ${I}_{2} = \int {\cos}^{2} x \mathrm{dx} = \frac{\sin x \cos x}{2} + \frac{1}{2} {I}_{0} = \frac{\sin x \cos x}{2} + \frac{1}{2} x$

$n = 3 \ddots$;${I}_{3} = \frac{\sin x {\cos}^{2} x}{2} + \frac{\cancel{2}}{\cancel{2}} \sin x$

$n = 4 \ddots$;${I}_{4} = \frac{\sin x {\cos}^{3} x}{2} + \frac{3}{2} {I}_{2} =$

$= \frac{\sin x {\cos}^{3} x}{2} + \frac{3}{2} \left(\frac{\sin x \cos x}{2} + \frac{1}{2} x\right) =$

$= \frac{\sin x {\cos}^{3} x}{2} + \frac{3 \sin x \cos x}{4} + \frac{3}{4} x$

And so on....

You can use other trigonometric identities to resume the final expresion...