Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} \), we use Trigonometric Identity, \({\tan ^2}x = {\sec ^2}x - 1\) then substitution method to integrate it.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Evaluation

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} \)

Let \(I = \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} \)

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}t \times {{\tan }^2}} tdt\)

We use \({\tan ^2}t = {\sec ^2}t - 1\)

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}t\left( {{{\sec }^2}t - 1} \right)} dt\\\therefore I &= \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t.{\sec ^2}tdt - \int\limits_0^{{\pi \mathord{\left/

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

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t.{\sec ^2}tdt - \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {\left( {{{\sec }^2}t - 1} \right)} dt\\ \Rightarrow I &= \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t.{\sec ^2}tdt - \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\sec }^2}tdt + \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {dt \to 1} } \end{aligned}\) \(\)

Step 2: Using Limits

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt\)

Let \(u = \tan t\)

\(\begin{aligned}{l}\frac{{du}}{{dt}} &= {\sec ^2}t\\ \Rightarrow du = {\sec ^2}tdt\end{aligned}\)

If \(t = 0 \Rightarrow u = \tan 0 \to 0\)

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

Therefore Lower Limit = 0 and Upper Limit = 1

\(\begin{aligned}{l}\therefore \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt &= \int\limits_0^1 {{u^2}.du} \\ &= \left( {\frac{{{u^{2 + 1}}}}{{2 + 1}}} \right)_0^1\\ &= \left( {\frac{{{u^3}}}{3}} \right)_0^1\\ &= \frac{1}{3}\left( {1 - 0} \right)\\\therefore \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^2}} t{\sec ^2}tdt &= \frac{1}{3} \to 2\end{aligned}\)

Step 3:Calculation

Finding the Integral of \(\int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\sec }^2}tdt} \)

\(\begin{aligned}{l}\int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\sec }^2}} tdt &= \left( {\tan t} \right)_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}}\\ &= \tan \frac{\pi }{4} - \tan 0\\ &= 1 - 0\\ &= 1\\\therefore \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\sec }^2}tdt &= 1 \to 3} \end{aligned}\)

Finding the value of \(\int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {dt} \)

\(\begin{aligned}{l}\int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {dt &= \left( t \right)_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}}} \\ &= \frac{\pi }{4} - 0\\\therefore \int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {dt &= \frac{\pi }{4}} \to 4\end{aligned}\)

Step 4: Substitution

Substituting the values of equations 2, 3 and 4

\(\begin{aligned}{l}\therefore I &= \frac{1}{3} - 1 + \frac{\pi }{4}\\ \Rightarrow I &= \frac{{4 - 12 + 3\pi }}{{12}}\\ \Rightarrow I &= \frac{{ - 8 + 3\pi }}{{12}} \Rightarrow I &= \frac{{3\pi - 8}}{{12}}\end{aligned}\)

Step 5: Replacing the equation

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} \)

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

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} = \frac{{3\pi - 8}}{{12}}\)

Hence the value of \(\int\limits_0^{{\pi \mathord{\left/

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

\kern-\nulldelimiterspace} 4}} {{{\tan }^4}tdt} \) is \(\frac{{3\pi - 8}}{{12}}\)

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Essential Calculus: Early Transcendentals

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Short Answer

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

Step by Step Solution

Step 1: Evaluation

Step 2: Using Limits

Step 3:Calculation

Step 4: Substitution

Step 5: Replacing the equation

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

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

Step by Step Solution

Step 1: Evaluation

Step 2: Using Limits

Step 3:Calculation

Step 4: Substitution

Step 5: Replacing the equation

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