91Ó°ÊÓ

u}\right| + C\), where \(n = 1\) and \(m = 2\). #tag_title# Step 5: Evaluate the transformed integral #tag_content# Using the table of integrals, we have: \(\frac{1}{3}\int\frac{\sec^3{\theta}}{\tan^2{\theta}}d\theta = \frac{1}{3}\left[\frac{1}{(1+1)} \log\left|\frac{\sin(1+1)\theta}{\sin{\theta}}\right| + C\right]= \frac{1}{6}\log\left|\frac{\sin{2\theta}}{\sin{\theta}}\right| + C\) #tag_title# Step 6: Substitute back for x #tag_content# Recall that our original substitution was \(x = 3\tan{\theta}\). To find \(\sin{2\theta}\) and \(\sin\theta\), we can use the following trigonometric identities: 1. \(\sin{2\theta} = 2\sin{\theta}\cos{\theta}\) 2. \(\tan\theta = \frac{x}{3}, \implies \theta = \arctan\left(\frac{x}{3}\right)\) Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) and the definition \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), we get \(\sin\theta = \frac{x}{\sqrt{x^2+9}}\) and \(\cos\theta = \frac{3}{\sqrt{x^2+9}}\). Now, we substitute these values for \(\sin{2\theta}\) and \(\sin{\theta}\): \(\frac{1}{6}\log\left|\frac{2\cdot\frac{x}{\sqrt{x^2+9}}\cdot\frac{3}{\sqrt{x^2+9}}}{\frac{x}{\sqrt{x^2+9}}}\right| + C = \frac{1}{6}\log\left|\frac{6}{\sqrt{x^2+9}}\right| + C\) #Answer# \(\frac{1}{6}\log\left|\frac{6}{\sqrt{x^2+9}}\right| + C\)