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

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