Question

Explain why, if $x=a\tan u$, then $\sqrt{x^2+a^2}=asecu$. Your explanation should include a discussion of domains and absolute values.

Short answer

Ans:

$$\begin{gathered} \sqrt{x^2+a^2}=asecu.......[x=a\tan u] \\ =\sqrt{a^2{\tan }^{2}u+a^2} \\ =asecu \end{gathered}$$

Step-by-step solution

Step 1. Given information:

$$\begin{gathered} x=a\tan u \\ \sqrt{x^2+a^2}=asecu \end{gathered}$$

Step 2. Solving the trigonometric substitution:

If $x=a\tan u$, there is no domain restriction on $x$, because $x=a\tan u$can be any real number. This means that trigonometric substitution $x=a\tan ux$ can be used even for integrands that do not have square root.

The domain of $x^2+a^2$ can be any value.

So, the integral involving $x^2+a^2$are well suited for trigonometric substitution with $x=a\tan u$.

Using the Pythagorean identity $a^2{\tan }^{2}u+a^2=a^2{\sec }^{2}u$.

$$\begin{gathered} \sqrt{x^2+a^2} \\ =\sqrt{a^2{\tan }^{2}u+a^2} \\ =\sqrt{a^{2}se{c}^{2}u} \\ =\left|asecu\right|[A.\sqrt{A^2}=\left|A\right|] \\ =asecu \end{gathered}$$