# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 31

$$\frac{{{{\sec }^4}4x}}{{16}} + C$$

#### Work Step by Step

\eqalign{ & \int {\tan 4x{{\sec }^4}4x} dx \cr & {\text{split }}{\sec ^4}x \cr & = \int {\tan 4x\sec 4x{{\sec }^3}4x} dx \cr & = \int {{{\sec }^3}4x\sec 4x\tan 4x} dx \cr & {\text{substitute }}u = \sec 4x,{\text{ }}du = \sec 4x\tan 4x\left( 4 \right)dx \cr & \frac{1}{4}du = \sec 4x\tan 4xdx \cr & = \int {{u^3}\left( {\frac{1}{4}du} \right)} \cr & = \frac{1}{4}\int {{u^3}du} \cr & {\text{find the antiderivative by the power rule}} \cr & = \frac{{{u^4}}}{{16}} + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr & = \frac{{{{\sec }^4}4x}}{{16}} + C \cr}

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