How to integrate: $\sin(x^2)$ How can we integrate $\sin\left(x^2\right)$?
$$f(x)=\int{\sin\left(x^2\right)}\ \mathrm dx$$
I almost lost about how to do it.
 A: There is no elementary solution. The integral is called Fresnel Integral. As you can see in the link, you can express it as a power series, but that's the most you can do.
A: There does not exist an elementary solution in a finite amount of terms. A non elementary solution would be in terms of the Fresnel S Integral, which would simply be:
$$\int \sin(x^2) dx=\sqrt{\frac{\pi}{2}}S(\sqrt{\frac{2}{\pi}} x)+C$$
However, you may find a power series for it, as mentioned by others, which is a series in terms of polynomials which converges towards the desired function. This series can be truncated (Remove all the terms starting from a particular term in the series) in order to obtain an approximation to your solution $\sqrt{\frac{\pi}{2}}S(\sqrt{\frac{2}{\pi}} x)+C$. To do so, use the Maclaurin series for $\sin(x)$.
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-
\frac{x^7}{7!}+...$$
Converting to sigma notation:
$$\sin(x)=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x)^{2n+1}}{(2n+1)!}$$
Evaluate the expansion of $\sin(x^2)$ by replacing $x$ with $x^2$.
$$\sin(x^2)=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x)^{4n+2}}{(2n+1)!}$$
Integrate both sides:
$$\int\sin(x^2) dx=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(x)^{4n+2}}{(2n+1)!} dx$$
Since you are integrating only in terms of $x$, this should be easy:
$$\int\sin(x^2) dx=\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{4n+3}}{(2n+1)!(4n+3)}+C$$
This may be represented as:
$$\int\sin(x^2) dx=\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320} - \frac{x^{15}}{75600} + ... + C$$
