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Found in: Page 340

Essential Calculus: Early Transcendentals

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

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

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

Hence,

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

We can see that both answer are same.