Question

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

Answer 1

`intcos^2(x^2)dx = x/2+(C(sqrt(2)x))/4 +C_0`

`C(2x)` denotes the Fresnel function evaluated at `2x` 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:
`intcos^2(x^2)dx`

Unfortunately:

`intcos(x^2)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(x) = intcos(x^2)dx`

By using the taylor series of `cos(x^2)` 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(x)` 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:

`intcos^2(x^2)dx=int1/2+1/2cos(2x^2)dx`

`=x/2+(C(sqrt(2)x))/4+C_0`

Where `C(sqrt(2)x)` 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(x^2)`.