How to find power series for Fresnel integral?

Question

How to get (step-by-step) from this

$$ C(x)=\int_{0}^{x} \cos \left(\frac{\pi t^{2}}{2}\right) d t $$

to this?

$$ C(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n}\left(\frac{\pi}{2}\right)^{2 n}}{(2 n) !(4 n+1)} x^{4 n+1} $$

Answer 1

\begin{align*} C(x)&=\color{red}{\int_0^x\cos\left(\frac{\pi t^2}2\right)\mathrm dt}=\int_0^x\sum_{n=0}^\infty(-1)^n\frac{\left(\frac{\pi t^2}2\right)^{2n}}{(2n)!}\mathrm dt=\sum_{n=0}^\infty(-1)^n\frac{\left(\frac\pi2\right)^{2n}}{(2n)!}\int_0^xt^{4n}\mathrm dt\\ &=\sum_{n=0}^\infty(-1)^n\frac{\left(\frac\pi2\right)^{2n}}{(2n)!}\left[\frac{t^{4n+1}}{4n+1}\right]_0^x=\color{red}{\sum_{n=0}^\infty(-1)^n\frac{\left(\frac\pi2\right)^{2n}}{(2n)!}\frac{x^{4n+1}}{4n+1}} \end{align*}

Answer 2

From $\cos u=\sum_{n\ge 0}\frac{(-1)^nu^{2n}}{(2n)!}$ we have $$C(x)=\sum_{n\ge0}\left(\frac{(-1)^n(\frac{\pi}{2})^{2n}}{(2n)!}\int_0^x t^{4n}dt\right)=\sum_{n\ge0}\frac{(-1)^n(\frac{\pi}{2})^{2n}x^{4n+1}}{(2n)!(4n+1)}.$$

Answer 3

$$\frac{\mathrm{d}}{\mathrm{d}x}C(x)=\cos{\left(\frac{\pi x^2}2\right)}=\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}\left(\frac{\pi x^2}2\right)^{2k}$$ Then integrate both sides term-by-term to get the final result. The constant of integration can be found by using $C(0)=0$.