Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.8 Logarithmic and Exponential - 6.8 Exercises: 32

Answer

$$ - \frac{{{4^{\cot x}}}}{{\ln 4}} + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{{4^{\cot x}}}}{{{{\sin }^2}x}}dx} \cr & {\text{trigonometric identity }}\csc x = \frac{1}{{\sin x}} \cr & \int {{4^{\cot x}}{{\csc }^2}xdx} \cr & {\text{substitute }}u = \cot x,{\text{ }}du = - {\csc ^2}xdx \cr & \int {{4^{\cot x}}{{\csc }^2}xdx} = - \int {{4^u}du} \cr & {\text{find the antiderivative}} \cr & {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & {\text{letting }}a = 4 \cr & = - \frac{{{4^u}}}{{\ln 4}} + C \cr & = - \frac{{{4^{\cot x}}}}{{\ln 4}} + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.