Why is $\int\frac{\sin (3\theta + \pi )+14}{15} \times \frac{\sin (3\theta - \pi )+14}{15}\,d\theta$ near double the area of $r=1$ with area $\pi?$ How is $$\int\frac{\sin (3\theta + \pi )+14}{15} \times \frac{\sin (3\theta - \pi )+14}{15}\,d\theta$$ almost double the area of $r=1$ with area $\pi?$
This is the graph of the function 
https://www.desmos.com/calculator/6q7fprhqak
This is plugged into integral calculator
This is it plugged into integral calculator
This must be a mistake. I am not sure why my integration is wrong though

In general $$\int_{0}^{2\pi} f(\theta)\,d\theta$$ is not the area of the curve, even when $f(\theta)>0$ for all $\theta$. This is because for $\Delta\theta$, the area is estimated by a triangle, not a rectangle. In particular, you have a triangle with lengths $f(\theta),f(\theta+\Delta \theta)$ and angle $\Delta\theta$. The triangle area is thus 
$$\frac{1}{2}f(\theta)f(\theta+\Delta\theta)\sin \Delta\theta.$$
Turns out, $\sin\Delta x\sim \Delta x$ is good enough. Thus, to get the area bounded by $f(\theta)$, you want:
$$\int_{0}^{2\pi} \frac{1}2f(\theta)^2\,d\theta$$
So you get a factor of $\frac{1}{2}$, plus you want to square your function.
It's a little trickier of $f(\theta)$ is negative for some values $\theta$.
See: http://tutorial.math.lamar.edu/Classes/CalcII/PolarArea.aspx
