Originally Posted by

**romsek** Start with

$\displaystyle \sin (x)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} x^{2 n-1}}{(2 n-1)!}$

$\displaystyle \sin \left(2 x^2\right)=\sin \left(\left(\sqrt{2} x\right)^2\right)$

$\displaystyle \sin \left(2 x^2\right)=\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}$

$\displaystyle \int \sin \left(2 x^2\right) \, dx$ = $\displaystyle \int \left(\sum _{n=1}^{\infty } \frac{(-1)^{n-1} \left(\sqrt{2} x\right)^{2 (2 n-1)}}{(2 n-1)!}\right) \, dx$ = $\displaystyle \sum _{n=1}^{\infty } \frac{(-1)^n \left(2^{2 n-1} x^{4 n-1}\right)}{(4 n-1) (2 n-1)!}$

The first 2 terms of this are

$\displaystyle \left\{-\frac{2 x^3}{3},\frac{4 x^7}{21}\right\}$

so you missed a negative sign. Not bad!