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 $0$?