Question

Prove part (b) of Theorem 5.18: For$x\in (-\infty ,\infty )\text{and}u\in (-\pi /2,\pi /2)\text{,}$the substitution $x=a\tan u$ gives $x^2+a^2=a\sec u$. Your proof should include a discussion of domains and a consideration of absolute values.

Short answer

It is proved.

Step-by-step solution

Step 1. Given information.

It is given$x\in (-\infty ,\infty ),u\in (-\pi /2,\pi /2)$

Step 2. Explanation.

For $x=a\tan u$

We know, From Pythagorean Theorem,

$$\begin{gathered} a^2{\tan }^{2}u+a^2=a^2{\sec }^{2}u \\ \sqrt{x^2+a^2} \\ =\sqrt{a^2{\tan }^{2}u+a^2} \\ =\sqrt{a^2{\sec }^{2}u} \\ =\left|a\sec u\right| \\ =a\sec u \end{gathered}$$

Hence proved.