I got this question from my calculus 2 textbook:
We are also given this known series:
Is there anyway to calculate arctan^2(x) using this table of already-solved series? I know we can manually derive it with a bunch of differentiation, but I'm assuming there's a faster way using the table that I just can't find.
Thanks.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#44f}{\arctan\pars{x}} & = \left.\int_{0}^{x}{\dd t \over 1 + t^{2}}\right\vert_{\, -1\ \leq\ x\ \leq\ 1}\,\,\, = \sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{x}t^{2n}\,\dd t \\[5mm] & = \bbx{\color{#44f}{\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over 2n + 1}\, x^{2n + 1}}} \\ & \end{align}