Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluate the integral\(\int\limits_0^{{\pi \mathord{\left/
{\vphantom {\pi 4}} \right.
\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

Evaluating the integral \(\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

We use identity \({\sec ^2}\theta = 1 + {\tan ^2}\theta \)

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

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Evaluation

Evaluating the integral\(\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

\(\begin{aligned}{l} \Rightarrow I &= \int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}\theta {{\sec }^2}\theta {{\sec }^2}\theta d\theta } \\ \Rightarrow I &= \int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}\theta (1 + {{\tan }^2}\theta ){{\sec }^2}\theta d\theta \to 1} \end{aligned}\)

Let \(u = \tan \theta \)

Differentiating with respect to\(\theta \)

\(\begin{aligned}{l}\frac{{du}}{{d\theta }} &= {\sec ^2}\theta \\ \Rightarrow du &= {\sec ^2}\theta d\theta \end{aligned}\)

Step 2: Integration Using Steps

Changing limits from \(\theta \)to \(u\)

If \(\theta = 0 \Rightarrow u = \tan o \to u = o\)

If \(\theta = \frac{\pi }{4} \Rightarrow u = \tan \frac{\pi }{4} \to u = 1\)

Upper Limit = 1 and Lower Limit = \(0\)

Step 3: Calculation

Substituting \(\tan \theta = u\) and \({\sec ^2}\theta d\theta = du\) in equation 1

And changing limits from \(0\)to \(\frac{\pi }{4}\)à\(0\)to \(1\)

Then equation 1 becomes

\(\begin{aligned}{l} \Rightarrow I &= \int\limits_0^1 {{u^4}(1 + {u^2})du} \\ \Rightarrow I &= \int\limits_{}^1 {} \end{aligned}\) \(\begin{aligned}{l}I &= \int\limits_0^1 {{u^4}\left( {1 + {u^2}} \right)} du\\ \Rightarrow I &= \int\limits_0^1 {{u^4}du + \int\limits_0^1 {{u^6}du} } \end{aligned}\)

We know that \(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}}\)

\(\begin{aligned}{l}I &= \left[ {\frac{{{u^{4 + 1}}}}{{4 + 1}}} \right]_0^1 + \left[ {\frac{{{u^{6 + 1}}}}{{6 + 1}}} \right]_0^1\\ &= \left[ {\frac{{{u^5}}}{5}} \right]_0^1 + \left[ {\frac{{{u^7}}}{7}} \right]_0^1\\ &= \frac{1}{5}\left( {1 - 0} \right) + \frac{1}{7}\left( {1 - 0} \right)\\ \Rightarrow I &= \frac{1}{5} + \frac{1}{7}\\ \Rightarrow I &= \frac{{7 + 5}}{{35}}\\ \Rightarrow I &= \frac{{12}}{{35}}\end{aligned}\)

Step 4: Finalizing the equation

Replacing 1 by \(\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

\(\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)\( = \frac{{12}}{{35}}\)

Hence, the value of Integral is

\(\int\limits_0^{{\pi \mathord{\left/

{\vphantom {\pi 4}} \right.

\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)\( = \frac{{12}}{{35}}\)

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Essential Calculus: Early Transcendentals

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Short Answer

Evaluate the integral\(\int\limits_0^{{\pi \mathord{\left/
{\vphantom {\pi 4}} \right.
\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

Step by Step Solution

Step 1: Evaluation

Step 2: Integration Using Steps

Step 3: Calculation

Step 4: Finalizing the equation

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Evaluate the integral\(\int\limits_0^{{\pi \mathord{\left/ {\vphantom {\pi 4}} \right. \kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)

Step by Step Solution

Step 1: Evaluation

Step 2: Integration Using Steps

Step 3: Calculation

Step 4: Finalizing the equation

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Evaluate the integral\(\int\limits_0^{{\pi \mathord{\left/
{\vphantom {\pi 4}} \right.
\kern-\nulldelimiterspace} 4}} {{{\sec }^4}\theta {{\tan }^4}\theta d\theta } \)