# How to integrate: $\sin(x^2)$ [duplicate]

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.

• This doesn't have an elementary anti-derivative. In general you shouldn't expect one. What's the context for this problem? Dec 28 '16 at 3:55
• If it seems hard or impossible, do it numerically. That's what I say. Dec 28 '16 at 3:56
• You can use the Taylor expansion for sin(x^2) and integrate that, but you won't be able to convert it to a closed form. Dec 28 '16 at 4:25

$$\int \sin(x^2) dx=\sqrt{\frac{\pi}{2}}S(\sqrt{\frac{2}{\pi}} x)+C$$
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: $$\int\sin(x^2) dx=\frac{x^3}{3} - \frac{x^7}{42} + \frac{x^{11}}{1320} - \frac{x^{15}}{75600} + ... + C$$