## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises: 46

#### Answer

$$- \frac{1}{6}cot6x + C$$

#### Work Step by Step

\eqalign{ & \int {{{\csc }^2}6x} dx \cr & {\text{use the formula for indefinite integrals of trigonometric functions}} \cr & \int {{{\csc }^2}ax} dx = - \frac{1}{a}cotax + C \cr & {\text{letting }}a = 6,{\text{ we have}} \cr & = - \frac{1}{6}cot6x + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( { - \frac{1}{6}cot6x + C} \right) \cr & {\text{ = }}\frac{d}{{dx}}\left( { - \frac{1}{6}cot6x} \right) + \frac{d}{{dx}}\left( C \right) \cr & {\text{ = }} - \frac{1}{6}\left( { - {{\csc }^2}6x} \right)\left( 6 \right) + 0 \cr & = {\csc ^2}6x \cr}

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