2. $int\frac{sqrt{9 - x^{2}}}{x^{2}}dx$

Explanation:

Step1: Use trig - substitution

Let (x = 3\sin\theta), then (dx=3\cos\theta d\theta). When (x = 3\sin\theta), (\sqrt{9 - x^{2}}=\sqrt{9 - 9\sin^{2}\theta}=3\cos\theta). The integral (\int\frac{\sqrt{9 - x^{2}}}{x^{2}}dx=\int\frac{3\cos\theta}{9\sin^{2}\theta}\cdot3\cos\theta d\theta=\int\frac{\cos^{2}\theta}{\sin^{2}\theta}d\theta).

Step2: Use trig - identity

Since (\cos^{2}\theta = 1-\sin^{2}\theta), the integral becomes (\int\frac{1 - \sin^{2}\theta}{\sin^{2}\theta}d\theta=\int(\csc^{2}\theta - 1)d\theta).

Step3: Integrate term - by - term

We know that (\int\csc^{2}\theta d\theta=-\cot\theta) and (\int 1d\theta=\theta + C). So (\int(\csc^{2}\theta - 1)d\theta=-\cot\theta-\theta + C).

Step4: Back - substitute

Since (x = 3\sin\theta), then (\sin\theta=\frac{x}{3}), (\theta=\arcsin(\frac{x}{3})) and (\cot\theta=\frac{\sqrt{9 - x^{2}}}{x}). So the integral (\int\frac{\sqrt{9 - x^{2}}}{x^{2}}dx=-\frac{\sqrt{9 - x^{2}}}{x}-\arcsin(\frac{x}{3})+C).

Answer:

(-\frac{\sqrt{9 - x^{2}}}{x}-\arcsin(\frac{x}{3})+C)