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?
Best Answer
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