Approximating an integral with taylor series

Question

I am working on the following homework problem:

"Assume that $\sin(x)$ equals its Maclaurin series for all $x$. Use the Maclaurin series for $\sin(5x^2)$ to evaluate the integral $$\int_0^{0.66} \sin(5x^2) dx$$ Your answer will be an infinite series. Use the first two terms to estimate its value."

Here's what I have come up with so far. $$\sin(x)=\sum_{n=0}^{\infty} \frac{(-1)^n(x^{2n+1})}{(2n+1)!}$$ Therefore $$\int_0^{0.66} \sin(5x)=\sum_{n=0}^{\infty} \frac{(-1)^n((5x)^{2n+2})}{(2n+1)!(2n+2)}\Bigg|_0^{0.66}$$ I then evaluate the summation for the first two n values after plugging evaluating the rest of the integral $$\frac{(-1)^0((5*0.66)^{2(0)+1})}{(2(0)+1)!}-\frac{(-1)^1((5*0.66)^{2(1)+1})}{(2(1)+1)!}$$ However this does not seem to get me the correct answer, which I calculated to be $\frac{40293}{80000}$. Where did I go wrong?

Answer 1

Since the Maclaurin series for $\sin x$ is $\sin x = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^nx^{2n+1}}{(2n+1)!}$, by replacing $x$ with $5x^2$, the Maclaurin series for $\sin(5x^2)$ is $\sin(5x^2) = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^n(5x^2)^{2n+1}}{(2n+1)!}$ $= \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^n5^{2n+1}x^{4n+2}}{(2n+1)!}$, which is different from what you got.

If you integrate this term by term, you should get the correct answer.