## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 14

#### Answer

$\sec^{-1} x+\dfrac{\sqrt{x^2 -1}}{x^2}+C$

#### Work Step by Step

Let us consider $x = \sec \theta$ and $dx= \sec\theta \tan \theta d \theta$ and $tan \theta =\sqrt {x^2-1}$ Now, the given integral becomes: $\int \dfrac{2 \sec \theta \tan \theta }{ \sec^3 \theta\tan \theta} d \theta =\int 2 \cos^2 \theta d \theta$ $2\cos^2 \theta =1+\cos 2 \theta$ Then, $\int 2 \cos^2 \theta d \theta= \int (1+cos 2 \theta ) d \theta$ Now, integrate and plug in $\sec \theta =x \implies \theta =\sec^{-1} x$ and $\cos \theta =\dfrac{1}{x}$ and $\sin \theta =\dfrac{\sqrt {x^2-1}}{x}$ : $\theta+\dfrac{1}{2} \sin (2 \theta)+C=\theta+\dfrac{1}{2} (2 \sin \theta \cos \theta)+C=\sec^{-1} x+\dfrac{\sqrt{x^2 -1}}{x^2}+C$

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