<s>[INST] Maclaurin Series 
Integrate using first three terms of appropriate series...
  $$\int_0^1 \sin x ~dx.$$

So I use
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}$$
for my three terms and if I integrate just that I get the answer which is $.3103$.
However the solution book is showing a negative is taken outside the integral then just sorta disappears.
Am I missing something here or is this a typo? (What follows is what is in the book):
$$\begin{align*}
\int_0^1\sin x^2\,dx &= \int_0^1\left(x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!}\right)\,dx\\
&= -\int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx\\
&= \left.\left(\frac{1}{3}x^3 - \frac{x^7}{42} + \frac{x^{11}}{1320}\right)\right|_0^1\\
&= \frac{1}{3} - \frac{1}{42} + \frac{1}{1320} = 0.3103.
\end{align*}$$
 [/INST] Yes, the minus sign is a typo and should not be there.
Note as well that you don't actually have equalities. Since
$$\sin x\approx x - \frac{x^3}{3!} + \frac{x^5}{5!},$$
and we don't have actual equality, we have that
$$\int_0^1\sin(x^2)\,dx \approx \int_0^1\left(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\right)\,dx.$$
Likewise, in the final line we actually have
$$\frac{1}{3} - \frac{1}{42} + \frac{1}{1320}\approx 0.3103,$$
not equality. 
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