The equation appears to involve integration with trigonometric functions. The integral shown is:

$\frac{\int_0^{2\pi} p \sin \theta \cos \theta \, d\theta}{\int_0^{2\pi} p \sin \theta \, d\theta}.$

To solve this:

Step 1: Simplify the Numerator

The numerator is: $\int_0^{2\pi} p \sin \theta \cos \theta \, d\theta.$

Using the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$, this becomes: $\int_0^{2\pi} \frac{p}{2} \sin(2\theta) \, d\theta.$

Since $\sin(2\theta)$ is a periodic function and integrates to zero over one full period $[0, 2\pi]$, the numerator simplifies to: $0.$

Step 2: Simplify the Denominator

The denominator is: $\int_0^{2\pi} p \sin \theta \, d\theta.$

Since $\sin(\theta)$ is also a periodic function and integrates to zero over $[0, 2\pi]$, the denominator also evaluates to: $0.$

Step 3: Evaluate the Fraction

The fraction is $\frac{0}{0}$, which is indeterminate. To resolve this, more information about $p$ is required—such as whether $p$ depends on $\theta$ or is constant.

Would you like to provide additional details about $p$, or clarify the problem? If $p$ is constant, let me know!

Here are five related questions to explore:

1. Does the parameter $p$ depend on $\theta$ or is it a constant?
2. Should this integral be solved for a specific domain other than $[0, 2\pi]$?
3. Are there alternative ways $p$ is defined in your context?
4. Is this part of a physics-related problem (e.g., torque or moment calculations)?
5. Can I explain trigonometric identities further for better clarity?

Tip: When simplifying integrals with periodic functions, always check the limits to see if symmetry helps simplify the work.