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**Failure** If you believe that $\displaystyle \sin(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$ then surely you have that $\displaystyle \sin(5x^2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}$.

Plugging this into your integral you get:

$\displaystyle \int_0^{0.64}\sin(5x^2)\, dx=\int_0^{0.64}\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx$

Next thing to do is to be bold enough to exhange $\displaystyle \int$ and $\displaystyle \sum$, like this:

$\displaystyle =\sum_{n=0}^\infty \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx\approx \sum_{n=0}^1 \int_0^{0.64}\frac{(-1)^n}{(2n+1)!}(5x^2)^{2n+1}\, dx=\ldots$