# How do you evaluate the integral of x*(cos(x^2))^5?

Mar 23, 2016

${\sin}^{5} \frac{{x}^{2}}{10} - {\sin}^{3} \frac{{x}^{2}}{3} + \sin \frac{{x}^{2}}{2} + C$

#### Explanation:

First, let

$u = {x}^{2} \text{ "=>" } \mathrm{du} = 2 x \mathrm{dx}$

So, we have

$\int x {\cos}^{5} \left({x}^{2}\right) \mathrm{dx}$

Rearrange the $x$ and $\mathrm{dx}$ and add a $2$ to achieve $2 x \mathrm{dx}$. To balance the $2$ inside the integral, multiply by $1 / 2$ outside of the integral.

$= \frac{1}{2} \int {\underbrace{{\cos}^{5} \left({x}^{2}\right)}}_{{\cos}^{5} \left(u\right)} \cdot {\underbrace{2 x \mathrm{dx}}}_{\mathrm{du}}$

Substituting, this equals

$= \frac{1}{2} \int {\cos}^{5} \left(u\right) \mathrm{du}$

Before we integrate this, we want to modify this so that we can have a $v$ and $\mathrm{dv}$ term both in the integral. We can do this by writing ${\cos}^{5} \left(u\right)$ as

${\cos}^{5} \left(u\right) = \cos \left(u\right) {\cos}^{4} \left(u\right) = \cos \left(u\right) {\left({\cos}^{2} \left(u\right)\right)}^{2} = \cos \left(u\right) {\left(1 - {\sin}^{2} \left(u\right)\right)}^{2}$

Giving us the integral

$= \frac{1}{2} \int \cos \left(u\right) {\left(1 - {\sin}^{2} \left(u\right)\right)}^{2} \mathrm{du}$

Now, we can substitute again:

$v = \sin \left(u\right) \text{ "=>" } \mathrm{dv} = \cos \left(u\right) \mathrm{du}$

Rearrange the integral:

$= \frac{1}{2} \int {\underbrace{{\left(1 - {\sin}^{2} \left(u\right)\right)}^{2}}}_{{\left(1 - {v}^{2}\right)}^{2}} \cdot {\underbrace{\cos \left(u\right) \mathrm{du}}}_{\mathrm{dv}}$

Which equals:

$= \frac{1}{2} \int {\left(1 - {v}^{2}\right)}^{2} \mathrm{dv}$

Distribute ${\left(1 - {v}^{2}\right)}^{2}$.

$= \frac{1}{2} \int \left(1 - 2 {v}^{2} + {v}^{4}\right) \mathrm{dv}$

Splitting up the integral, we obtain:

$= \frac{1}{2} \int {v}^{4} \mathrm{dv} - 2 \left(\frac{1}{2}\right) \int {v}^{2} \mathrm{dv} + \frac{1}{2} \int 1 \mathrm{dv}$

Using the rule:

$\int {v}^{n} \mathrm{dv} = {v}^{n + 1} / \left(n + 1\right) + C$

Yields, after integrating term by term:

$= \frac{1}{2} \left({v}^{5} / 5\right) - {v}^{3} / 3 + \frac{1}{2} \left(v\right) + C$

$= {v}^{5} / 10 - {v}^{3} / 3 + \frac{v}{2} + C$

Using $v = \sin \left(u\right)$:

$= {\sin}^{5} \frac{u}{10} - {\sin}^{3} \frac{u}{3} + \sin \frac{u}{2} + C$

Now using $u = {x}^{2}$:

$= {\sin}^{5} \frac{{x}^{2}}{10} - {\sin}^{3} \frac{{x}^{2}}{3} + \sin \frac{{x}^{2}}{2} + C$