Chapter 5.1, Problem 91E

To determine

To express: The given algebraic expression as a trigonometric function of $\theta$ .

Answer to Problem 91E

The expression $\sqrt{{\left(1+x^2\right)}^3}$ is equal to ${\sec }^3\theta$ , when $x=\tan \theta$ .

Explanation of Solution

Given information :

The given algebraic expression:

  $\sqrt{{\left(1+x^2\right)}^3}$

The given trigonometric substitution:

  $x=\tan \theta$

Where $0<\theta <\pi /2$

Calculation :

As per problem,

The given expression:

  $\sqrt{{\left(1+x^2\right)}^3}$

Let $x=\tan \theta$

  $\begin{array}{l}\sqrt{{\left(1+x^2\right)}^3}=\sqrt{{\left(1+{\left(\tan \theta \right)}^2\right)}^3}\text{ [Substitute }\tan \theta \text{ for }x]\\ =\sqrt{{\left(1+{\tan }^2\theta \right)}^3}\text{ [Rule of exponents}\\ =\sqrt{{\left({\sec }^2\theta \right)}^3}\text{ [Pythagorean identity: }{\sec }^2\theta \text{=}1+{\text{tan}}^2\theta \text{]}\\ =\sqrt{{\sec }^6\theta }\text{ [Write the cube]}\\ ={\sec }^3\theta \text{ [sec}\theta \text{>0 for 0<}\theta \text{<}\pi /2\text{]}\end{array}$

Hence, the expression $\sqrt{{\left(1+x^2\right)}^3}$ is equal to ${\sec }^3\theta$ ,when $x=\tan \theta$ .