# How do you integrate int x^2 sin^2 x^2 dx  using integration by parts?

Apr 21, 2018

$\int {x}^{2} {\sin}^{2} \left({x}^{2}\right) \mathrm{dx} = \frac{1}{6} {x}^{3} - \frac{1}{4} x \sin \left(2 {x}^{2}\right) - \frac{1}{4} \int \sin \left(2 {x}^{2}\right) \mathrm{dx}$

#### Explanation:

First, let's use the identity ${\sin}^{2} \left(\alpha\right) = \frac{1}{2} \left(1 - \cos \left(2 \alpha\right)\right)$:

$\int {x}^{2} {\sin}^{2} \left({x}^{2}\right) \mathrm{dx} = \int {x}^{2} \left[\frac{1}{2} \left(1 - \cos \left(2 {x}^{2}\right)\right)\right] \mathrm{dx}$

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

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

The first integral is very easy:

$= \frac{1}{6} {x}^{3} - \int {x}^{2} \cos \left(2 {x}^{2}\right) \mathrm{dx}$

The second integral is a little trickier. Let's now try to do this by parts.

When I see the cosine function which has the argument ${x}^{2}$, I expect for the function to be multiplied by $x$ , of degree $1$, based on the rough idea that $\frac{d}{\mathrm{dx}} \sin \left({x}^{2}\right) = 2 x \cos \left({x}^{2}\right)$.

So, with this in mind, let's let $\mathrm{dv} = x \cos \left(2 {x}^{2}\right) \mathrm{dx}$ and $u = x$, which is what remains in the integrand.

Finding $\mathrm{du}$ is simple: $\mathrm{du} = \mathrm{dx}$. Finding $v$ takes a little thinking. Let's integrate $\mathrm{dv}$ with the substitution $t = 2 {x}^{2} \implies \mathrm{dt} = 4 x \mathrm{dx}$.

$v = \int x \cos \left(2 {x}^{2}\right) \mathrm{dx} = \frac{1}{4} \int \cos \left(2 {x}^{2}\right) \left(4 x \mathrm{dx}\right) = \frac{1}{4} \int \cos \left(t\right) \mathrm{dt} = \frac{1}{4} \sin \left(t\right) = \frac{1}{4} \sin \left(2 {x}^{2}\right)$

Then, using $\int u \mathrm{dv} = u v - \int v \mathrm{du}$, the original integral simplifies to become:

$= \frac{1}{6} {x}^{3} - \left[x \left(\frac{1}{4} \sin \left(2 {x}^{2}\right)\right) - \int \frac{1}{4} \sin \left(2 {x}^{2}\right) \mathrm{dx}\right]$

$= \frac{1}{6} {x}^{3} - \frac{1}{4} x \sin \left(2 {x}^{2}\right) - \frac{1}{4} \int \sin \left(2 {x}^{2}\right) \mathrm{dx}$

Integrals in the form that we see remaining, those resembling ${\int}_{0}^{x} \sin \left({z}^{2}\right) \mathrm{dz}$ or ${\int}_{0}^{x} \cos \left({z}^{2}\right) \mathrm{dz}$, don't have very common closed forms, so this is where I'd stop.

Maybe you're at a higher level of calculus than I am, in which case, I refer you to the following page on Fresnel integrals: https://en.wikipedia.org/wiki/Fresnel_integral