Questions: Evaluate [ int fracsqrtx^2-9x^3 d x ] First, write the appropriate trigonometric substitution. Use (theta) for the substitution variable. [ x= ]

Solution Steps

We start with the integral \[ \int \frac{\sqrt{x^{2}-9}}{x^{3}} d x. \] Using the substitution \(x = 3\sec(\theta)\), we can express \(\sqrt{x^2 - 9}\) as \(\sqrt{(3\sec(\theta))^2 - 9} = \sqrt{9\sec^2(\theta) - 9} = 3\sqrt{\sec^2(\theta) - 1} = 3\tan(\theta)\).

Next, we compute the differential \(dx\): \[ dx = 3\sec(\theta)\tan(\theta) d\theta. \] Substituting \(x\) and \(dx\) into the integral gives: \[ \int \frac{3\tan(\theta)}{(3\sec(\theta))^3} \cdot 3\sec(\theta)\tan(\theta) d\theta = \int \frac{3\tan^2(\theta)}{27\sec^2(\theta)} \cdot 3\sec(\theta)\tan(\theta) d\theta = \int \frac{3\tan^2(\theta)}{9\sec(\theta)} d\theta. \]

The integral simplifies to: \[ \int \frac{\tan^2(\theta)}{3\sec(\theta)} d\theta = \frac{1}{3} \int \tan^2(\theta) \cos(\theta) d\theta. \] Using the identity \(\tan^2(\theta) = \sec^2(\theta) - 1\), we can further simplify the integral: \[ \frac{1}{3} \int (\sec^2(\theta) - 1) \cos(\theta) d\theta. \]

The integral can be evaluated as: \[ \frac{1}{3} \left( \int \cos(\theta) d\theta - \int \cos(\theta) d\theta \right) = \frac{1}{3} \left( \sin(\theta) - \theta \right) + C. \]

Finally, we substitute back to \(x\) using \(\theta = \sec^{-1}\left(\frac{x}{3}\right)\) and \(\sin(\theta) = \frac{\sqrt{x^2 - 9}}{x}\): \[ \frac{1}{3} \left( \frac{\sqrt{x^2 - 9}}{x} - \sec^{-1}\left(\frac{x}{3}\right) \right) + C. \]

Final Answer