Answer
$r(\theta)=8\theta+cot\theta-2\pi-1$
Work Step by Step
Step 1. Given $r'(\theta)=8-csc^2\theta$, we can find a function $f(\theta)=8\theta+cot\theta$ such that $f'(\theta)=8-csc^2\theta$.
Step 2. Based on Corollary 2, we have $r(\theta)=f(\theta)+C=8\theta+cot\theta+C$ where $C$ is a constant.
Step 3. Using the condition that $P(\frac{\pi}{4},0)$ is on the curve, we have $0=8\pi/4+cot\pi/4+C$ which gives $C=-2\pi-1$
Step 4. Thus, we get the function as $r(\theta)=8\theta+cot\theta-2\pi-1$
