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The following question uses Maclaurin series to investigate approximations of mathematical constants and the accuracies of such approximations.

By using integration by parts, show that $\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x \mathrm{~d} x=\frac{\pi}{6 \sqrt{3}}-\frac{1}{2} \ln \frac{4}{3}$ .

$\quad u=\arctan x$ and $\frac{\mathrm{d} v}{\mathrm{~d} x}=1$ (or equivalent)

\int_{0}^{\frac{1}{\sqrt{3}}} \arctan x \mathrm{~d} x=[x \arctan x]_{0}^{\frac{1}{\sqrt{3}}}-\int_{0}^{\frac{1}{\sqrt{3}}} \frac{x}{1+x^{2}} \mathrm{~d} x

Note: Award A1 for each term.

\begin{aligned} & =[x \arctan x]_{0}^{\frac{1}{\sqrt{3}}}-\frac{1}{2}\left[\ln \left(1+x^{2}\right)\right]_{0}^{\frac{1}{\sqrt{3}}}\left(=\frac{1}{\sqrt{3}} \arctan \frac{1}{\sqrt{3}}-\frac{1}{2} \ln \frac{4}{3}\right) \\ & \quad=\frac{\pi}{6 \sqrt{3}}-\frac{1}{2} \ln \frac{4}{3} \end{aligned}

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