Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - Review Exercises - Page 579: 48

Answer

$$\int {{{\tan }^n}xdx} = \frac{1}{{n - 1}}{\tan ^{n - 1}}x - \int {{{\tan }^{n - 2}}xdx} $$

Work Step by Step

$$\eqalign{ & \int {{{\tan }^n}xdx} \cr & {\text{Recall that }}{a^{m + n}} = {a^m}{a^n},{\text{ then}} \cr & \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2 + 2}}xdx} \cr & \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x{{\tan }^2}xdx} \cr & {\text{Rewrite }}{\tan ^2}x{\text{ using the identity ta}}{{\text{n}}^2}x = {\sec ^2}x - 1 \cr & \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} \cr & \int {{{\tan }^n}xdx} = \int {{{\tan }^{n - 2}}x{{\sec }^2}xdx} - \int {{{\tan }^{n - 2}}xdx} \cr & {\text{Integrate }}\int {{{\tan }^{n - 2}}x{{\sec }^2}xdx} {\text{ by using the power rule}} \cr & \int {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 2 + 1}}x}}{{n - 2 + 1}} - \int {{{\tan }^{n - 2}}xdx} \cr & {\text{Simplifying}} \cr & \int {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} \cr & \int {{{\tan }^n}xdx} = \frac{1}{{n - 1}}{\tan ^{n - 1}}x - \int {{{\tan }^{n - 2}}xdx} \cr} $$
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