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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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    Patrick: Yes. ${}$2012-08-14

1 Answers 1

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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 $[0, \frac{1}{2}]$.