# 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

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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.

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Yeah, it adds up "0s" Doesn't from $3\pi/2$ to $2\pi$ it sweeps everything else? –  sidht Jul 19 '12 at 4:55
The part from $3\pi/2$ to $2\pi$ is part of the "real" curve, since $\cos\theta$ is $\ge 0$ there. –  André Nicolas Jul 19 '12 at 4:56
What do you mean "real" part? –  sidht Jul 19 '12 at 4:59
I mean the curve. The part with $r$ negative is not part of the curve. Actually, under a popular interpretation, $r$ negative is OK, but if you get that you reflect across the origin. With that interprtation you really do traverse the circle twice, once would be $0$ to $\pi$. –  André Nicolas Jul 19 '12 at 5:05
@jak: Have you tried tracing out the curve in polar coordinates as you go through the values of $\theta$ within your integration domain? You're basically drawing over the curve anew, and the integration won't catch that. –  Ｊ. Ｍ. Jul 19 '12 at 5:09

$r = \cos \theta$ 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.

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