Question

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

Answer 1

`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(alpha)=1/2(1-cos(2alpha))`:

`intx^2sin^2(x^2)dx=intx^2[1/2(1-cos(2x^2))]dx`

`=1/2int(x^2-x^2cos(2x^2))dx`

`=1/2intx^2dx-intx^2cos(2x^2)dx`

The first integral is very easy:

`=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`, I expect for the function to be multiplied by `x` , of degree `1`, based on the rough idea that `d/dxsin(x^2)=2xcos(x^2)`.

So, with this in mind, let's let `dv=xcos(2x^2)dx` and `u=x`, which is what remains in the integrand.

Finding `du` is simple: `du=dx`. Finding `v` takes a little thinking. Let's integrate `dv` with the substitution `t=2x^2=>dt=4xdx`.

`v=intxcos(2x^2)dx=1/4intcos(2x^2)(4xdx)=1/4intcos(t)dt=1/4sin(t)=1/4sin(2x^2)`

Then, using `intudv=uv-intvdu`, the original integral simplifies to become:

`=1/6x^3-[x(1/4sin(2x^2))-int1/4sin(2x^2)dx]`

`=1/6x^3-1/4xsin(2x^2)-1/4intsin(2x^2)dx`

Integrals in the form that we see remaining, those resembling `int_0^xsin(z^2)dz` or `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