Integrating definite integrals in terms of area Lately, I've been trying to come up with tricks to solve integrals quickly.
So let's say I have $$\int_{0}^{2\pi} \cos^2 \theta d\theta$$
Now if I were to look at this integral in polar coordiantes, I get
$$\frac{1}{2}\int_{0}^{2\pi} \cos^2 \theta d\theta$$
The integrand is a circle in polar coordinates $r = 2(1/2)\cos\theta$ with radius $1/2$
So the integral $$\frac{1}{2}\int_{0}^{2\pi} \cos^2 \theta d\theta = \frac{\pi}{2}$$
But this doesn't make sense to me because the area should be $\pi (1/2)^2 = \pi/4$
What's going on? I am trying to extend this idea to $$\int_{0}^{2\pi} \sin^2 \theta d\theta$$ and linear combinations of sine and cosines squared
 A: $r = \cos \theta$ describes a circle for, say, $-\pi/2 \le r \le \pi/2$ (where $\cos \theta > 0$).
As $\theta$ goes from $0$ to $2 \pi$, you go around the circle twice.
A: Your integral does not "know" that $r$ cannot be negative.  So from $\pi/2$ to $3\pi/2$ it cheerfully keeps "adding  up" $\cos^2 \theta$, not realizing there is no curve there. Effectively the integral traverses the circle twice.  
It really should know that the integration should be done from $0$ to $\pi/2$, and from $3\pi/2$ to $2\pi$. Or equivalently that it should be integrating $\frac{1}{2}f^2(\theta)$, where $f(\theta)=0$ on $(\pi/2,3\pi/2)$, and $f(\theta)=\cos\theta$ elsewhere on $[0,2\pi]$. But nobody told it. 
Remark: There is not universal agreement that the point that has polar coordinates address $(r,\theta)$ makes no sense if $r \lt 0$. One common interpretation in that case is that you graph $(r,\theta)$ and then reflect the result across the origin. That interpretation sometimes produces nicer pictures. 
If we take that interpretation, then $r=\cos\theta$ sweeps out our circle as $\theta$ travels from $0$ to $\pi$, so integrating from $0$ to $\pi$ gives the area. 
