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Q12E

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Essential Calculus: Early Transcendentals

Found in: Page 306

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Evaluate the indefinite integral\(\int {{{\sec }^2}2} \theta d\theta \).

The indefinite integral value of the given equation is\(\int {{{\sec }^2}} 2\theta d\theta = \frac{1}{2}\tan \theta + c\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1 Definition of the indefinite integral

An integral that has no upper bound and no lower bound is known as indefinite integral.

Step 2: Expand the number of variables in the equation.

Given: \(I = \int {{{\sec }^2}} 2\theta d\theta \)

Known value \(\int {{{\sec }^2}} axdx = \frac{1}{a}\tan ax + c\)

Step 3: Evaluate the equation.

Put\(a = 2,x = \theta \)

So, \(I = \frac{1}{2}\tan 2\theta + c\)

Therefore, the indefinite integral value of the given equation is\(\int {{{\sec }^2}} 2\theta d\theta = \frac{1}{2}\tan \theta + c\).

Most popular questions for Math Textbooks

Evaluate the indefinite integral\(\int {\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}} dx\).

Evaluate the integral by making the given substitution.

\(\int {{{\cos }^3}} \theta \sin \theta d\theta ,\;\;\;u = \cos \theta \)

Use Property 8 to estimate the value of the integral.

\(\int_1^2 {\frac{1}{x}} dx\)

Evaluate the integral by making the given substitution.

\(\int {{{\rm{x}}^{\rm{2}}}} \sqrt {{{\rm{x}}^{\rm{3}}}{\rm{ + 1}}} {\rm{dx,}}\;\;\;{\rm{u = }}{{\rm{x}}^{\rm{3}}}{\rm{ + 1}}\).

Calculate the area of the region that lies under the curve and above the x-axis.

\({\rm{y = 1 - }}{{\rm{x}}^{\rm{2}}}\)

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