# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 47

$$\ln \left| {6 + 3\tan t} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{3{{\sec }^2}t}}{{6 + 3\tan t}}} dt \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = 6 + 3\tan t,{\text{ so that }}du = 3{\sec ^2}tdt \cr & dt = \frac{{du}}{{3{{\sec }^2}t}} \cr & {\text{write the integral in terms of }}u \cr & \int {\frac{{3{{\sec }^2}t}}{{6 + 3\tan t}}} dt = \int {\frac{{3{{\sec }^2}t}}{u}} \left( {\frac{{du}}{{3{{\sec }^2}t}}} \right) \cr & = \int {\frac{1}{u}} \left( {\frac{{du}}{1}} \right) \cr & = \int {\frac{1}{u}} du \cr & {\text{integrate }} \cr & = \ln \left| u \right| + C \cr & {\text{write in terms of }}t{\text{; replace }}6 + 3\tan t{\text{ for }}u \cr & = \ln \left| {6 + 3\tan t} \right| + C \cr}

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