# Maclaurin series approximation

So, I've been running through some of these problems in my text book and was fine until I came across a Maclaurin series of a definite integral.

$$\int_0^{1/2}\tan^{-1}(2x^2) dx$$

from the table in the book I know that

$$\int\tan^{-1}(2x^2) dx = \sum_0^\infty(-1)^k\frac{(2x^2)^{2k+1}}{2k+1}$$

Is it just as simple as plugging $2x^2$ in for $x$.

Am I supposed to integrate this first or just evaluate the sum as $1/2$ and $0$?

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Well, for typesettings, use tan^{-1} instead of tan^-1. – Frenzy Li Aug 14 '12 at 4:45
Thanks! Any clue why (2k+1) as an exponent shows up strange? – Patrick M Aug 14 '12 at 4:46
What you can do is to start with the series for the arctangent, replace the variable with $2x^2$, and then integrate termwise... – J. M. Aug 14 '12 at 4:46
@PatrickM It's because you could have fit all the equation code into a single block: $L.H.S = R.H.S$. :) – Frenzy Li Aug 14 '12 at 4:48
@Frenz, no, it's because OP did not enclose his exponents within braces. Compare $x^-1$ (x^-1) with $x^{-1}$ (x^{-1}). – J. M. Aug 14 '12 at 4:52

The series $\arctan z = \sum_{k=0}^\infty (-1)^n \frac{z^{2k+1}}{2k+1}$ is valid for $|z| <1$. If $x \in [0, \frac{1}{2}]$, then $2x^2 \in [0,\frac{1}{2}]$, so we can integrate the series with $z=2 x^2$ term by term.
So plug $z=2x^2$ into the series, and integrate over $[0, \frac{1}{2}]$.