## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 44

#### Answer

$$\frac{{2{{\sec }^{3/2}}x}}{3} + C$$

#### Work Step by Step

\eqalign{ & \int {\tan x} {\sec ^{3/2}}xdx \cr & {\text{split exponent of }}{\sec ^{3/2}}x \cr & = \int {\tan x} {\sec ^{1/2}}x\sec xdx \cr & = \int {{{\sec }^{1/2}}x} \sec x\tan xdx \cr & {\text{substitute }}u = \sec x,{\text{ }}du = \sec x\tan xdx \cr & = \int {{u^{1/2}}} du \cr & {\text{find the antiderivative by the power rule}} \cr & = \frac{{{u^{3/2}}}}{{3/2}} + C \cr & = \frac{{2{u^{3/2}}}}{3} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \sec x \cr & = \frac{{2{{\sec }^{3/2}}x}}{3} + C \cr}

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