Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.1 Integration by Parts - 7.1 Exercises - Page 517: 53


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

Work Step by Step

Given $$\int \tan ^{n} x d x$$ Since \begin{align*} \int \tan ^{n} x d x&=\int \tan ^{n-2} x \tan ^{2} x d x\\ &=\int \tan ^{n-2} x\left(\sec ^{2} x-1\right) d x\\ &=\int \tan ^{n-2} x \sec ^{2} x d x-\int \tan ^{n-2} x d x \end{align*} Let $u=\tan x \ \Rightarrow \ \ du =\sec^2 x dx$, then \begin{align*} \int \tan ^{n-2} x \sec ^{2} x d x&=\int u ^{n-2} d u\\ &=\frac{u^{n-1}}{n-1}\\ &=\frac{\tan^{n-1}x}{n-1}+c \end{align*} Hence $$\int \tan ^{n} x d x =\frac{\tan^{n-1}x}{n-1}-\int \tan ^{n-2} x d x$$
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