# Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 37

$$\frac{1}{3}{\left( {\tan x} \right)^3} + C$$

#### Work Step by Step

\eqalign{ & \int {{{\sec }^2}x} {\tan ^2}xdx \cr & = \int {{{\tan }^2}x{{\sec }^2}x} dx \cr & {\text{Integrate by substitution method}} \cr & {\text{Let }}u = \tan x,\,\,\,\,du = {\sec ^2}xdx,\,\,\,\,dx = \frac{{du}}{{{{\sec }^2}x}} \cr & {\text{Then}}{\text{,}} \cr & \int {{{\tan }^2}x{{\sec }^2}xdx} = \int {{u^2}{{\sec }^2}x\left( {\frac{{du}}{{{{\sec }^2}x}}} \right)} \cr & {\text{Cancel common factor }}{\sec ^2}x \cr & = \int {{u^2}} du \cr & {\text{integrating}} \cr & = \frac{1}{3}{u^3} + C \cr & {\text{write in terms of }}x \cr & = \frac{1}{3}{\left( {\tan x} \right)^3} + C \cr}

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