Answer
$0.0059$
Work Step by Step
$arctan(x)=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)}$
$\int_{0}^{1/2}arctan(x)dx=\int_{0}^{1/2}x^{3}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)}dx$
$\int_{0}^{1/2}arctan(x)dx=\int_{0}^{1/2}\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+4}}{(2n+1)}dx$
Then
$=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+5}}{(2n+5)(2n+1)}$
$=\frac{0.5^{2}}{5}-\frac{0.5^{7}}{21}+\frac{0.5^{9}}{45}$
$\approx 0.0059$