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Q18E

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Found in: Page 326

### Essential Calculus: Early Transcendentals

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

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# Evaluate the Integral: $$\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta$$

To evaluate the Integral$$\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta$$we will use identity $${\sec ^2}\theta = 1 + {\tan ^2}\theta$$ then use the substitution method to get the result.

$$\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta = \frac{1}{3}{\tan ^3}\theta + \frac{1}{5}{\tan ^5}\theta + c$$

See the step by step solution

## Step 1: Splitting the fourth power of the secant function

Let I = $$\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta$$

First, we will write $${\sec ^4}\theta = {\sec ^2}\theta {\sec ^2}\theta$$ then use $${\sec ^2}\theta = 1 + {\tan ^2}\theta$$

## Step 2: Using trigonometric identity

\begin{aligned}{l}I &= \int {{{\tan }^2}} \theta {\sec ^2}\theta {\sec ^2}\theta d\theta \\ &= \int {{{\tan }^2}} \theta {\sec ^2}\theta (1 + {\tan ^2}\theta )d\theta \end{aligned}

## Step 3: Substitution method

Now we will use the substitution method

Put $$\tan \theta = x( = ){\sec ^2}\theta d\theta = dx$$

$$I = \int {({x^2}} + {x^4})dx$$

Split the integral

\begin{aligned}{l}I = \int {{x^2}} dx + \int {{x^4}} dx\\I = \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} + c\end{aligned}

## Step 4 : Re-Substitute the value of $$x = \tan \theta$$

$$I = \frac{{{{\tan }^3}\theta }}{3} + \frac{{{{\tan }^5}\theta }}{5} + c$$

Hence, the value of integral:

$$\int {{{\tan }^2}} \theta {\sec ^4}\theta d\theta = \frac{1}{3}{\tan ^3}\theta + \frac{1}{5}{\tan ^5}\theta + c$$

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