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Essential Calculus: Early Transcendentals
Found in: Page 340
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Use a computer algebra system to evaluate the integral. Compare the answer with the result using tables. If the answer is not the same, Show that they are equivalent.

\(\int {{{\sec }^4}xdx} \)

Using any computer algebra system:-

\(\int {{{\sec }^4}xdx = \frac{{{{\tan }^3}x}}{3} + \tan x + c} \)

See the step by step solution

Step by Step Solution

Step(1):- calculating the values using any computer software

Using an integral calculator, we get

\(\int {{{\sec }^4}xdx = \frac{{{{\tan }^3}x}}{3} + \tan x} \)

Step(2):- Calculating the value of integral using table

\(\begin{aligned}{l}I = \int {{{\sec }^4}xdx} \\I = \int {{{\sec }^2}x.{{\sec }^2}xdx} \end{aligned}\) \((1 + {\tan ^2}x = {\sec ^2}x)\)

\(\begin{aligned}{l}I &= \int {{{\sec }^2}x.(1 + {{\tan }^2}x)} dx\\I &= \int {{{\sec }^2}x + } {\sec ^2}x{\tan ^2}xdx\\I &= \int {{{\sec }^2}xdx + \int {{{\sec }^2}x{{\tan }^2}xdx} } \\I &= \int {{{\sec }^2}xdx} + \int {{{\sec }^2}x{{(\tan x)}^2}dx} \end{aligned}\)

We know, \(\int {{{\sec }^2}xdx = \tan x} \)

Also, \(\int {f{{\left( x \right)}^n}.{f^'}\left( x \right)dx = \frac{{{{(f\left( x \right))}^{n + 1}}}}{{n + 1}}} + C\)

\(\int {{{(\tan x)}^2}.{{\sec }^2}xdx = \frac{{{{\tan }^3}x}}{3}} + C\)


\(\int {{{\sec }^4}x = \tan x + \frac{{{{\tan }^3}x}}{3} + C} \)

We can see that both answer are same.

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