# How do you find the integral of f(x)=cos^2(x^2) using integration by parts?

Jan 11, 2018

$\int {\cos}^{2} \left({x}^{2}\right) \mathrm{dx} = \frac{x}{2} + \frac{C \left(\sqrt{2} x\right)}{4} + {C}_{0}$

$C \left(2 x\right)$ denotes the Fresnel function evaluated at $2 x$ and ${C}_{0}$ denotes the constant of integration. This may be expressed in terms of the power series:

#### Explanation:

I could take a stab at the solution but it involves a different approach to integration by parts.

We have:
$\int {\cos}^{2} \left({x}^{2}\right) \mathrm{dx}$

Unfortunately:

$\int \cos \left({x}^{2}\right) \mathrm{dx}$

does not have a nice solution that can be expressed in terms of elementary functions. There is however some hope, there are a pair of special functions called the Fresnel functions (which are usually covered a bit beyond beginner calculus though it has very important applications in optics) and one of them is defined as follows:

$C \left(x\right) = \int \cos \left({x}^{2}\right) \mathrm{dx}$

By using the taylor series of $\cos \left({x}^{2}\right)$ we can see that:

C(x) =intsum_(n=0)^oo((-1)^nx^(4n))/((2n)!)dx=sum_(n=0)^oo((-1)^nx^(4n+1))/((4n+1)(2n)!)

The function looks like this: For simplicity we will simply denote the Fresnel function as $C \left(x\right)$ from now on and if you wish to find out more then follow the link:
https://en.wikipedia.org/wiki/Fresnel_integral

So using the appropriate trig identity:

$\int {\cos}^{2} \left({x}^{2}\right) \mathrm{dx} = \int \frac{1}{2} + \frac{1}{2} \cos \left(2 {x}^{2}\right) \mathrm{dx}$

$= \frac{x}{2} + \frac{C \left(\sqrt{2} x\right)}{4} + {C}_{0}$

Where $C \left(\sqrt{2} x\right)$ is the Fresnel function and ${C}_{0}$ is the constant of integration.

We can stop here or write this in terms of the power series from above:

=x/2+1/4sum_(n=0)^oo((-1)^n(sqrt(2)x)^(4n+1))/((4n+1)(2n)!)+C_0

Unfortunately, for reasons stated above, this is about as good as you can get to an integral for ${\cos}^{2} \left({x}^{2}\right)$.