Solving $y'=\cos x \int \cos x^2 dx$

This seems a very hard ODE that I couldn't solve.

It is from Zorich Mathematical Analysis so I guess there must be a somewhat-enlightening solution.

Though my half-attempt is here:

$$y'(x)=\cos x\int \cos x^2 dx$$$$dy=\cos x(\int \cos x^2 dx) dx$$$$y=\int \cos x(\int \cos x^2 dx)dx+C$$

Editted after 4 hours:

By using trigonometric series for $\cos x^2$ and $\cos x$

We get $$y(x)=\int \cos x\int \cos x^2 dx= \int \cos x\sum_{k=0}^\infty\frac{(-1)^kx^{4k+1}}{(2k)!(4k+1)} dx$$$$y(x)=\sum_{k=0}^\infty \frac{(-1)^k \int x^{4k+1}\cos x dx}{(2k)!(4k+1)}$$

By adopting a power series approach again,

$$y(x)=\sum_{i,j=0}^\infty \frac{(-1)^j x^{2j+4k+2}}{(2k)!(2j)!(4k+1)(2j+4k+2)}$$

Now, my question is, is there a closed form for this? Or even, a better approach?