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 $\pi/12$.


u &=& \arctan(3 x)\\
dv &=& dx\;.


du &=&\displaystyle \frac{3}{9 x^2 + 1} dx\\
v &=& x\;.


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

we get

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


\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}


\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}

If you prefer to jump to the next problem, click on Next Problem below.

[Next Problem] [Matrix Algebra]
[Trigonometry] [Calculus]
[Geometry] [Algebra]
[Differential Equations] [Complex Variables]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Mohamed A. Khamsi

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour