Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 24


$$\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{t}{8\left(4+9 t^{2}\right)}+C$$

Work Step by Step

Given $$\int \frac{d t}{\left(9 t^{2}+4\right)^{2}}$$ Let $$3t=2\tan u\ \ \ \ \ \to\ \ \ \ 3dt= 2\sec^2 udu $$ Then \begin{align*} \int \frac{d t}{\left(9 t^{2}+4\right)^{2}}&=\frac{2}{3} \int \frac{\sec^2 udu }{\left(4\tan ^2 u+4\right)^{2}}\\ &= \frac{1}{24} \int \frac{\sec^2 udu }{\sec^{4}u}\\ &= \frac{1}{24}\int \cos^2 udu\\ &= \frac{1}{48}\int (1+\cos 2u)du\\ &=\frac{1}{48}\left[u+\frac{\sin 2 u}{2}\right]+C\\ &=\frac{1}{48} u+\frac{1}{48} \sin u \cos u+C\\ & = \frac{1}{48} u+\frac{1}{48} \sin u \cos u+C \\ &=\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{1}{48} \frac{3 t}{\sqrt{4+9 t^{2}}} \frac{2}{\sqrt{4+9 t^{2}}}+C\\ &=\frac{1}{48} \tan ^{-1}\left(\frac{3 t}{2}\right)+\frac{t}{8\left(4+9 t^{2}\right)}+C \end{align*}
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