# Chapter 7 - Principles Of Integral Evaluation - 7.4 Trigonometric Substitutions - Exercises Set 7.4 - Page 513: 4

$$- \frac{{\sqrt {9 - {x^2}} }}{{9x}} + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{dx}}{{{x^2}\sqrt {9 - {x^2}} }}} \cr & {\text{substitute }}x = 3\sin \theta ,{\text{ }}dx = 3\cos \theta d\theta \cr & = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - {{\left( {3\sin \theta } \right)}^2}} }}} \cr & = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\sqrt {9 - 9{{\sin }^2}\theta } }}} \cr & = \int {\frac{{3\cos \theta d\theta }}{{{{\left( {3\sin \theta } \right)}^2}\left( 3 \right)\sqrt {1 - {{\sin }^2}\theta } }}} \cr & = \frac{1}{9}\int {\frac{{\cos \theta d\theta }}{{{{\sin }^2}\theta \sqrt {{{\cos }^2}\theta } }}} \cr & = \frac{1}{9}\int {\frac{1}{{{{\sin }^2}\theta }}} d\theta \cr & = \frac{1}{9}\int {{{\csc }^2}\theta } d\theta \cr & {\text{find antiderivative}} \cr & = \frac{1}{9}\left( { - \cot \theta } \right) + C \cr & = - \frac{1}{9}\cot \theta + C \cr & {\text{write in terms of }}x \cr & = - \frac{1}{9}\left( {\frac{{\sqrt {9 - {x^2}} }}{x}} \right) + C \cr & = - \frac{{\sqrt {9 - {x^2}} }}{{9x}} + C \cr}

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