# 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

• First of all $\sin(3\theta+\pi)=\sin(3\theta-\pi)$. Is this what you meant? – Thomas Andrews Aug 16 '17 at 23:24
• Whatever gets multiplied with $theta$ and also the phase shift doesn't seem to matter. When you integrate from r:0-1 ; $theta$:0-2$pi$ I get $(131/75)*pi$ – Austin Nguyen Aug 16 '17 at 23:27
• Austin, write \theta with a dollar sign at the start and end to create $\theta$. Do the same for Pi with \pi instead. A "multiplication sign" can be expressed as \times with a dollar sign also at the start and end. – George N. Missailidis Aug 16 '17 at 23:30

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 $\theta$.