Problem 19 - Essential Calculus

Question

Find the integral. $$ \int \frac{1}{\sqrt{x^{2}-9}} d x $$

Step-by-Step Solution

Answer

The integral is \(ln| \frac{x+\sqrt{x^{2}-9}}{3}| + C\).

Step 1: Substitution

Choose the trigonometric substitution \(x = a \sec \theta = 3 \sec \theta\). This leads to \(dx = 3 \sec \theta \tan \theta d\theta\) and \(\sqrt{x^2 - a^2} = \sqrt{9 \sec^2 \theta - 9} = 3 |\tan \theta|\). For this substitution, x > 0 leads to \(\tan \theta > 0\).

Step 2: Substitute in the Integral

Substitute all the obtained values into the integral: \(\int \frac{1}{\sqrt{x^{2}-9}} dx = \int \frac{1}{3 |\tan \theta|} \cdot 3 \sec \theta \tan \theta d\theta = \int \sec \theta d\theta\)

Step 3: Simplify the Integral

The integral of \(\sec \theta\) is a standard integral whose solution is given by \(ln|\sec \theta + \tan \theta|\). Thus: \(\int \sec \theta d\theta = ln|sec\theta + tan\theta|\)

Step 4: Back-substitute x

Now, replace \(\theta\) with the original variable \(x = 3\sec\theta\). Thus: \(ln|sec\theta + tan\theta| = ln|\frac{x}{3} + \frac{\sqrt{x^{2}-9}}{3}| = ln| \frac{x+\sqrt{x^{2}-9}}{3}|.\), as \(\theta = \sec^{-1}(\frac{x}{3})\). Rearranging gives \(ln| \frac{x+\sqrt{x^{2}-9}}{3}| + C\), where C is a constant of integration.