Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluating the integral \(\int {{{\tan }^5}} x{\sec ^3}xdx\)

Evaluating the integral \(\int {{{\tan }^5}} x{\sec ^3}xdxjaae\) we use the identity \({\tan ^2}x = {\sec ^2}x - 1\)

If the power of \(\tan x\) is odd, save a factor of \(\sec x\tan x\) and use \({\tan ^2}x = {\sec ^2}x - 1\)to express the remaining factors in term of \(\sec x\). Then substitute \(u = \sec x\)

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Evaluation

Evaluating the Integral

\({\tan ^2}x = {\sec ^2}x - 1\)= \(\int {{{\tan }^4}x{{\sec }^2}} x.\sec x\tan xdx\)

= \(\int {{{\left( {{{\tan }^2}x} \right)}^2}} {\sec ^2}x.\sec xtanxdx\)

Use Trigonometric Identity \({\tan ^2}x = {\sec ^2}x - 1\)

\(\therefore \int {{{\tan }^5}} x{\sec ^2}xdx = {\int {\left( {{{\sec }^2}x - 1} \right)} ^2}{\sec ^2}x.\sec x\tan xdx \to 1\)

Let \(u = \sec x\)

Differentiating with respect to \(x\)

\(\begin{aligned}{l}\frac{{du}}{{dx}} &= \sec x\tan x\\ \Rightarrow du &= \sec x\tan xdx\end{aligned}\)

Substituting \(\sec x = u\) and \(\sec x\tan xdx = du\) in equation 1

Then equation 1 becomes

Therefore, \(\int {{{\tan }^5}} x{\sec ^3}xdx = {\int {\left( {{u^2} - 1} \right)} ^2}{u^2}.du \to 2\)

Use Identity \({\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\)

So, \({\left( {{u^2} - 1} \right)^2} = {u^4} + 1 - 2{u^2}\)

Substitute the above value in equation 2

\(\begin{aligned}{l}\int {{{\tan }^5}} x{\sec ^3}xdx &= \int {\left( {{u^4} + 1 - 2{u^2}} \right)} {u^2}du\\ &= \int {\left( {{u^6} + {u^2} - 2{u^4}} \right)} du\\\therefore \int {{{\tan }^5}} x{\sec ^3}xdx &= \int {{u^6}} du + \int {{u^2}du - 2\int {{u^4}du} } \\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{u^{6 + 1}}}}{{6 + 1}} + \frac{{{u^{2 + 1}}}}{{2 + 1}} - 2.\frac{{{u^{4 + 1}}}}{{4 + 1}} + c\\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{u^7}}}{7} + \frac{{{u^3}}}{3} - 2.\frac{{{u^5}}}{5} + c \to 3\end{aligned}\)

Step 2: Substitution

Substituting the value \(u = \sec x\)in equation 3

Thus, \(\begin{aligned}{l}\int {{{\tan }^5}} x{\sec ^3}xdx &= \frac{{{{\left( {\sec x} \right)}^7}}}{7} + \frac{{{{\left( {\sec x} \right)}^3}}}{3} - \frac{{2{{\left( {\sec x} \right)}^5}}}{5} + c\\ \Rightarrow \int {{{\tan }^5}} x{\sec ^3}xdx = \frac{{{{\sec }^7}x}}{7} + \frac{{{{\sec }^3}x}}{3} - \frac{{2{{\sec }^5}x}}{5} + c\end{aligned}\)

Hence the value of \(\int {{{\tan }^5}} x{\sec ^3}xdx\)

\(\frac{{{{\sec }^7}x}}{7} + \frac{{{{\sec }^3}x}}{3} - \frac{{2{{\sec }^5}x}}{5} + c\)

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Essential Calculus: Early Transcendentals

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Short Answer

Evaluating the integral \(\int {{{\tan }^5}} x{\sec ^3}xdx\)

Step by Step Solution

Step 1: Evaluation

Step 2: Substitution

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Essential Calculus: Early Transcendentals

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Short Answer

Evaluating the integral \(\int {{{\tan }^5}} x{\sec ^3}xdx\)

Step by Step Solution

Step 1: Evaluation

Step 2: Substitution

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