Evaluate Definite Integral to desired accuracy Evaluate $$\int_0^{1/2}x^3\arctan(x)\,dx$$
My work so far:
$x^3\arctan(x) = \sum_{n=0}^\infty(-1)^n \dfrac{x^{2n+4}}{2n+1}$
$$\int_0^{1/2}x^3\arctan(x)\,dx = \sum_{n=0}^\infty \frac{(-1)^{n+1}(1/2)^{2n+4}}{2n+1}$$
The desired accuracy needed is to four decimal places, that is were I am stuck.
 A: Hint: You might consider first integrating parts:
$$\begin{align}
\int_{0}^{1/2}x^3\arctan{(x)}\,\mathrm{d}x
&=\frac14x^4\arctan{(x)}\bigg{|}_{0}^{1/2}-\frac14\int_{0}^{1/2}\frac{x^4}{1+x^2}\,\mathrm{d}x\\
&=\frac{1}{64}\arctan{\left(\frac12\right)}-\frac14\int_{0}^{1/2}\frac{x^4}{1+x^2}\,\mathrm{d}x\\
&=\frac{1}{64}\arctan{\left(\frac12\right)}-\frac14\int_{0}^{1/2}\left(x^2-1+\frac{1}{1+x^2}\right)\,\mathrm{d}x.
\end{align}$$
A: You wrote correctly $$x^3\arctan(x) = \sum_{n=0}^\infty(-1)^n \dfrac{x^{2n+4}}{2n+1}$$ but you forgot to integrate with respect to $x$. If you do it $$\int_0^{1/2}x^3\arctan(x)\,dx = \sum_{n=0}^\infty \frac{(-1)^{n}(1/2)^{2n+5}}{(2n+1)(2n+5)}$$
As said in the comments, you could start searching for which value of $n$ $$\frac{(1/2)^{2n+5}}{(2n+1)(2n+5)} \leq 5\times 10^{-5}$$ that is to say $$(2n+1)(2n+5)2^{2n+5}  \geq 20000$$ Probably, plotting the function after taking logarithms of both sides will make life easier and $n=2$  satisfies the condition ($23040$ for the lhs).
The other solution, also said in comments, is to start adding the terms and so, you have for the integral $$\frac{1}{160}-\frac{1}{2688}+\frac{1}{23040}=\frac{191}{32256} \simeq 0.00592138$$
If you take the time of computing exactly the integral, as David H suggested in his answer, you probably obtained $$\int x^3\arctan{(x)}\,\mathrm{d}x=\frac{1}{4} x^4 \tan ^{-1}(x)-\frac{x^3}{12}+\frac{x}{4}-\frac{1}{4} \tan ^{-1}(x)$$ and then $$\int_{0}^{1/2}x^3\arctan{(x)}\,\mathrm{d}x=\frac{1}{192} \left(22-45 \tan ^{-1}\left(\frac{1}{2}\right)\right) \simeq 0.00591592$$
