# How does one integrate $\int \cos(x^2) dx$

How does one integrate $\int \cos(x^2) dx$?

I have thought about using standard techniques of 'integration by parts' and 'partial fractions' but neither of them work. I tried plugging it into Wolfram Alpha and I got $\sqrt{\frac{\pi}{2}} C {\sqrt{\frac{2}{\pi}}} +$ a constant.

I am confused as to what the $C$ is supposed to mean and how to evaluate this intergral.

• At the bottom right of the output cell on WolframAlpha it says that "$C(x)$ is the Fresnel C integral". If you hover your mouse over that text, several links (such as this one) pop up describing it. The integral cannot be expressed as a finite combination of elementary functions. – Mark McClure Mar 21 '15 at 6:42
• $$\int_{-\infty}^\infty\sin\big(x^2\big)~dx ~=~ \int_{-\infty}^\infty\cos\big(x^2\big)~dx~=~\sqrt{\frac\pi2}$$ and $$\int_{-\infty}^\infty e^{-x^2}~dx~=~\sqrt\pi$$ The two identities above, related to Fresnel and Gaussian integrals, are linked to each other by way of Euler's formula. – Lucian Mar 21 '15 at 8:19
• reference.wolfram.com/language/ref/FresnelC.html – Aditya Hase Mar 21 '15 at 16:41

There's no elementary integral for this function i.e. it can't be solved in closed form. This is because there's no closed form anti-derivative of cos($x^2$). You'll have to use some kind of numerical method to solve it.The most straightforward method of solving it is to use the Taylor expansion of cosine replacing x with $x^2$:
$$\cos(x^2) = \sum_{n=0}^\infty (-1)^n \frac{(x^2)^{2n}}{(2n)!}$$