# Arc Length of a Cardioid $r=2\sin{\theta}-2$; Issue with Bounds of Integration

Good morning. I am trying to find the arc length of the cardioid $r=2\sin{\theta}-2$. After plugging in $r$ and $\frac{dr}{d\theta}$ into the arc length formula we get

$$s=\int_\alpha^\beta\sqrt{(2\sin\theta-2)^2+(2\cos\theta)^2}d\theta=2\sqrt{2}\int_\alpha^\beta{\sqrt{1-\sin\theta}} d\theta$$

It appears from a graphing calculator that the bounds of integration cover the principal angles $0\le\theta<2\pi$. Using an identity

$$\sqrt{1-\sin\theta}=\sqrt{\frac{1-\sin^2\theta}{1+\sin\theta}}=\frac{\cos\theta}{\sqrt{1+\sin\theta}}$$

This gives me the integral

$$s=2\sqrt{2}\int_0^{2\pi}{\frac{\cos\theta}{\sqrt{1+\sin\theta}}} d\theta$$

So now to solve I make a $u$-substitution. Let $u=1+\sin\theta$. Then $du=\cos\theta d\theta$. But if $\theta=0$, $u=1$. And if $\theta=2\pi$, then also $u=1$. This creates a situation where the bounds of integration are the same, which makes a calculation of $0$ for the final answer. So where did I go wrong?

• Which formula do you used? $$\int \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$? – Dr. Sonnhard Graubner Apr 29 '18 at 13:20
• Yes. Under the radical i got $$(2\sin\theta-2)^2+(2\cos\theta)^2$$ and when I expand, I get $$4\sin^2\theta-8\sin\theta+4+4\cos^2\theta$$ which reduces to $8-8\sin\theta$ – Lalaloopsy Apr 29 '18 at 13:22

HINT...You can, by consideration of the symmetry of the polar graph in the $y$ axis, rewrite this integral as $$s=\color{red}{2\times}2\sqrt{2}\int_{\color{red}{-\frac{\pi}{2}}}^{\color{red}{\frac{\pi}{2}}}{\frac{\cos\theta}{\sqrt{1+\sin\theta}}} d\theta$$
Now the bounds for $u$ are straightforward
Hint: use $$\sin(\theta)=\frac{2t}{1+t^2}$$ and $$d\theta=\frac{2}{1+t^2}dt$$
• After looking at the graph, I think it might be easier to exploit the symmetry of the cardiod and multiply the integral by 2 and use bounds $\pi/2$ to $3\pi/2$? – Lalaloopsy Apr 29 '18 at 13:36
It's much easier if you do\begin{align}1-\sin\theta&=\sin\left(\frac\pi2\right)-\sin\theta\\&=\sin\left(\left(\frac\pi4+\frac\theta2\right)+\left(\frac\pi4-\frac\theta2\right)\right)-\sin\left(\left(\frac\pi4+\frac\theta2\right)-\left(\frac\pi4-\frac\theta2\right)\right)\\&=2\sin\left(\frac\pi4+\frac\theta2\right)\cos\left(\frac\pi4-\frac\theta2\right)\\&=2\sin\left(\frac\pi4+\frac\theta2\right)\cos\left(\frac\pi2-\left(\frac\pi4+\frac\theta2\right)\right)\\&=2\sin^2\left(\frac\pi4+\frac\theta2\right),\end{align}from wich you deduce that$$\sqrt{1-\sin\theta}=\sqrt2\left|\sin\left(\frac\pi4+\frac\theta2\right)\right|.$$