Problems on Techniques of Integration

The Arctangent function is one of these functions whose derivative is very nice (rational function). So it is a good idea to use the integration by parts technique in which $\arctan(x)$ will be differentiated. We may also use the antiderivative of the function $\arctan(x)$ and evaluate it at the points $0$ and $\pi/12$.

Set

\begin{displaymath}\left\{\begin{array}{lll}
u &=& \arctan(3 x)\\
dv &=& dx\;.
\end{array}\right.\end{displaymath}

Then

\begin{displaymath}\left\{\begin{array}{lll}
du &=&\displaystyle \frac{3}{9 x^2 + 1} dx\\
v &=& x\;.
\end{array}\right.\end{displaymath}

Since

\begin{displaymath}\int_a^b u dv = [u v]_a^b - \int_a^b v du\;,\end{displaymath}

we get

\begin{displaymath}\begin{array}{lll}
\displaystyle \int_0^{\pi/12} \arctan(3 x)...
... \frac{1}{6}\Big[\ln(9x^2 + 1)\Big]_0^{\pi/12}\cdot
\end{array}\end{displaymath}

Since

\begin{displaymath}\Big[x \arctan(3 x)\Big]_0^{\pi/12} = \frac{\pi}{12} \arctan\left(\frac{\pi}{4}\right) = \frac{\pi}{12}\end{displaymath}

and

\begin{displaymath}\Big[\ln(9x^2 + 1)\Big]_0^{\pi/12} = \ln\left(\frac{9 \pi^2}{144} + 1 \right) = \ln\left(\frac{ \pi^2}{16} + 1 \right)\end{displaymath}

we get

\begin{displaymath}\int_0^{\pi/12} \arctan(3 x) dx = \frac{\pi}{12} - \frac{1}{6}\ln\left(\frac{ \pi^2}{16} + 1 \right)\;.\end{displaymath}


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