Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 491: 59

Answer

$\int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 1}x}}}{{n - 1}} - \int {{{\tan }^{n - 2}}x} dx,{\text{ }}\left( {n \ne 1} \right)$

Work Step by Step

$$\eqalign{ & \int {{{\tan }^n}x} dx \cr & {\text{Split the integrand using the property of exponents }}{a^m}{a^n} = {a^{m + n}} \cr & \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2 + 2}}x} dx \cr & \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x{{\tan }^2}x} dx \cr & {\text{Apply the pythagorean identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr & \int {{{\tan }^n}x} dx = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)} dx \cr & \int {{{\tan }^n}x} dx = \int {\left( {{{\tan }^{n - 2}}x{{\sec }^2}x} \right)} dx - \int {{{\tan }^{n - 2}}x} dx \cr & {\text{Integrating}}{\text{, let }}u = \tan x,{\text{ }}du = {\sec ^2}xdx,{\text{ then}} \cr & \int {{{\tan }^n}x} dx = \overbrace {\int {\left( {{{\tan }^{n - 2}}x{{\sec }^2}x} \right)} dx}^{\int {{u^{n - 2}}} du} - \int {{{\tan }^{n - 2}}x} dx \cr & \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 2 + 1}x}}}{{n - 2 + 1}} - \int {{{\tan }^{n - 2}}x} dx \cr & {\text{Simplifying}} \cr & \int {{{\tan }^n}x} dx = \frac{{{{\tan }^{n - 1}x}}}{{n - 1}} - \int {{{\tan }^{n - 2}}x} dx,{\text{ }}\left( {n \ne 1} \right) \cr} $$
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