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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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    Well, for typesettings, use `tan^{-1}` instead of `tan^-1`.2012-08-14
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    Thanks! Any clue why (2k+1) as an exponent shows up strange?2012-08-14
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    What you can do is to start with the series for the arctangent, replace the variable with $2x^2$, and then integrate termwise...2012-08-14
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    @PatrickM It's because you could have fit all the equation code into a single block: `$ L.H.S = R.H.S $`. :)2012-08-14
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    @Frenz, no, it's because OP did not enclose his exponents within braces. Compare $x^-1$ (`x^-1`) with $x^{-1}$ (`x^{-1}`).2012-08-14
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    @J.M. Oh, turns out you're right, but using `$ sth. $ = $ sth. $ $ factor $` is as weird, I suppose.2012-08-14
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    $\int_0^.5 \dfrac{(2x^2)^1}{1!} + \dfrac{(2x^2)^3}{3!} + \dfrac{(2x^2)^5}{5!} + \dfrac{(2x^2)^7}{7!} + ... dx$ Leaves me with... $\dfrac{2x^3}{3*1!} + \dfrac{8x^7}{7*3!} + \dfrac{32x^11}{11*5!} + \dfrac{128x^15}{15*7!} + ... ]_0^.5$2012-08-14
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    Didn't expect it to submit that when I inserted a line break. From there I can plug in x=1/2 and evaluate it and that would result in my approximate sum?2012-08-14
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    Patrick: Yes. ${}$2012-08-14

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