How to integrate $x^2\sin(x^2)$? $$\int x^2\sin(x^2)dx$$
Methods I know of: Reversing the chain, substitution, parts. Any help greatly appreciated.
Edit: Note that the integral is not $x^2\ast(\sin x)^2$
 A: I'm afraid that this is one integral where there is no nice form for the antiderivative. There are many ways to estimate the antiderivative. The simplest is to use the power series representation of $\sin(x)$.
The Maclaurin series for $\sin(x)$ was known as far back as the 14th century by  Madhava of Sangamagrama. This series is often covered about halfway through a Calculus 2 course. It is given by $$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1}=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots.$$
Replacing $x^2$ for $x$ gives:
$$\sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{4n+2}$$
and multiplying by $x^2$ gives:
$$x^2 \sin(x^2) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{4n+4}.$$
Now if we want to antidifferentiate this quantity we find:
$$\int x^2 \sin(x^2) dx = C+ \sum_{n=0}^\infty \frac{(-1)^n}{(4n+5)(2n+1)!} x^{4n+5}.$$
A: Fresnel's integrals are defined as:
$S(x) = \int\limits_{0}^{x} \sin\left(\frac{\pi}{2}\alpha^2\right) ~ d\alpha$
$C(x) = \int\limits_{0}^{x} \cos\left(\frac{\pi}{2}\alpha^2\right) ~ d\alpha$
Sometimes the $\frac{\pi}{2}$ is left out, but I chose to include it.
If you are a student in high school calculus, you should have learned the second fundamental theorem, which states that if $F(x)$ can be defined as
$$F(x) = \int\limits_{c}^{x} f(\alpha) ~ d\alpha$$
Then $F(x)$ is an antiderivative of $f(x)$.
Perhaps your teacher wanted you to realize that the integral is not expressible by elementary functions, and define a new function (namely the Fresnel functions) using the fundamental theorem to solve it. You don't need to have been previously aware of the Fresnel integral to do this. It is also possible that your teacher assigned this knowing you cannot do it, so that they can later make the point that integration is much harder than differentiation.
A: Here's a way to prove that this integral comes down to the non-elementary Fresnel integral: integrate by parts.  Taking $u=x, dv=x\sin(x^2) dx$ we have
$du=dx$ (trivially) and $v=\int x\sin(x^2) dx = -\frac12\cos(x^2)$, so we get
$$\int x^2\sin(x^2)\ dx = \int x\cdot x\sin(x^2)\ dx = -\frac12x\cos(x^2)+\int\frac12\cos(x^2)\ dx$$
Since we've found a relation in terms of elementary functions between $\int x^2\sin(x^2)\ dx$ and $\int \cos(x^2)\ dx$, then if one were expressible in terms of elementary functions then the other would have to be; but it is widely known (as other answers have mentioned) that $\int \cos(x^2)\ dx$ cannot be expressed in terms of elementary functions, so neither can $\int x^2\sin(x^2)\ dx$.
