How do you integrate $\int x^{2} {sin}^{2} x^{2} d x$ $int x^2 sin^2 x^2 dx$ using integration by parts?

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How do you integrate $\int x^{2} {sin}^{2} x^{2} d x$ $int x^2 sin^2 x^2 dx$ using integration by parts?

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Answer 1 · 2018-04-21T01:18:40

$\int x^{2} {sin}^{2} (x^{2}) d x = \frac{1}{6} x^{3} - \frac{1}{4} x sin (2 x^{2}) - \frac{1}{4} \int sin (2 x^{2}) d x$ $intx^2sin^2(x^2)dx=1/6x^3-1/4xsin(2x^2)-1/4intsin(2x^2)dx$

Explanation:

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

$\int x^{2} {sin}^{2} (x^{2}) d x = \int x^{2} [\frac{1}{2} (1 - cos (2 x^{2}))] d x$ $intx^2sin^2(x^2)dx=intx^2[1/2(1-cos(2x^2))]dx$

$= \frac{1}{2} \int (x^{2} - x^{2} cos (2 x^{2})) d x$ $=1/2int(x^2-x^2cos(2x^2))dx$

$= \frac{1}{2} \int x^{2} d x - \int x^{2} cos (2 x^{2}) d x$ $=1/2intx^2dx-intx^2cos(2x^2)dx$

The first integral is very easy:

$= \frac{1}{6} x^{3} - \int x^{2} cos (2 x^{2}) d x$ $=1/6x^3-intx^2cos(2x^2)dx$

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 $x^{2}$ $x^2$ , I expect for the function to be multiplied by $x$ $x$ , of degree $1$ $1$ , based on the rough idea that $\frac{d}{d x} sin (x^{2}) = 2 x cos (x^{2})$ $d/dxsin(x^2)=2xcos(x^2)$ .

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

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

$v = \int x cos (2 x^{2}) d x = \frac{1}{4} \int cos (2 x^{2}) (4 x d x) = \frac{1}{4} \int cos (t) d t = \frac{1}{4} sin (t) = \frac{1}{4} sin (2 x^{2})$ $v=intxcos(2x^2)dx=1/4intcos(2x^2)(4xdx)=1/4intcos(t)dt=1/4sin(t)=1/4sin(2x^2)$

Then, using $\int u d v = u v - \int v d u$ $intudv=uv-intvdu$ , the original integral simplifies to become:

$= \frac{1}{6} x^{3} - [x (\frac{1}{4} sin (2 x^{2})) - \int \frac{1}{4} sin (2 x^{2}) d x]$ $=1/6x^3-[x(1/4sin(2x^2))-int1/4sin(2x^2)dx]$

$= \frac{1}{6} x^{3} - \frac{1}{4} x sin (2 x^{2}) - \frac{1}{4} \int sin (2 x^{2}) d x$ $=1/6x^3-1/4xsin(2x^2)-1/4intsin(2x^2)dx$

Integrals in the form that we see remaining, those resembling $\int_{0}^{x} sin (z^{2}) d z$ $int_0^xsin(z^2)dz$ or $\int_{0}^{x} cos (z^{2}) d z$ $int_0^xcos(z^2)dz$ , 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

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How do you integrate $\int x^{2} {sin}^{2} x^{2} d x$ $int x^2 sin^2 x^2 dx$ using integration by parts?

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