evaluate $\\int \\frac{\\sqrt{x^{2}-36}}{x^{3}} d x$.

Explanation:

Step1: Use trigonometric substitution

Let (x = 6\sec\theta), then (dx=6\sec\theta\tan\theta d\theta). And (\sqrt{x^{2}-36}=\sqrt{36\sec^{2}\theta - 36}=6\tan\theta), (x^{3}=216\sec^{3}\theta). The integral (\int\frac{\sqrt{x^{2}-36}}{x^{3}}dx=\int\frac{6\tan\theta}{216\sec^{3}\theta}\cdot6\sec\theta\tan\theta d\theta) Simplify it: (\frac{1}{6}\int\frac{\tan^{2}\theta}{\sec^{2}\theta}d\theta=\frac{1}{6}\int\sin^{2}\theta d\theta)

Step2: Use double - angle formula

Since (\sin^{2}\theta=\frac{1 - \cos2\theta}{2}), then (\frac{1}{6}\int\sin^{2}\theta d\theta=\frac{1}{12}\int(1-\cos2\theta)d\theta) Integrate term - by - term: (\frac{1}{12}(\theta-\frac{1}{2}\sin2\theta)+C) Using the double - angle formula (\sin2\theta = 2\sin\theta\cos\theta), we get (\frac{1}{12}\theta-\frac{1}{12}\sin\theta\cos\theta+C)

Step3: Back - substitute

Since (x = 6\sec\theta), then (\sec\theta=\frac{x}{6}), (\cos\theta=\frac{6}{x}), (\sin\theta=\frac{\sqrt{x^{2}-36}}{x}), and (\theta=\text{arcsec}(\frac{x}{6})) The integral becomes (\frac{1}{12}\text{arcsec}(\frac{x}{6})-\frac{\sqrt{x^{2}-36}}{12x^{2}}+C)

Answer:

(\frac{1}{12}\text{arcsec}(\frac{x}{6})-\frac{\sqrt{x^{2}-36}}{12x^{2}}+C)