## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 50

#### Answer

$$- \frac{1}{3}\csc 3t + C$$

#### Work Step by Step

\eqalign{ & \int {{{\cot }^2}3t\sec 3t} dt \cr & {\text{Use the identities }}\cot x = \frac{{\cos x}}{{\sin x}},\,\,\,\,\sec x = \frac{1}{{\cos x}} \cr & = \int {{{\left( {\frac{{\cos 3t}}{{\sin 3t}}} \right)}^2}\left( {\frac{1}{{\cos 3t}}} \right)} dt \cr & {\text{Simplifying}} \cr & = \int {\frac{{{{\cos }^2}3t}}{{{{\sin }^2}3t}}\left( {\frac{1}{{\cos 3t}}} \right)} dt \cr & = \int {\frac{{\cos 3t}}{{{{\sin }^2}3t}}} dt \cr & \cr & {\text{Use the substitution method}}{\text{, }}u = \sin 3t,\,\,\,du = 3\cos 3tdt \cr & = \int {\frac{{\cos 3t}}{{{u^2}}}\left( {\frac{{du}}{{3\cos 3t}}} \right)} \cr & = \frac{1}{3}\int {\frac{{du}}{{{u^2}}}} \cr & {\text{Integrate}} \cr & = - \frac{1}{{3u}} + C \cr & {\text{Write in terms of }}t \cr & = - \frac{1}{{3\sin 3t}} + C \cr & = - \frac{1}{3}\csc 3t + C \cr}

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