# Find the arc length of the polar curve $r=1+\cos(t)$.

How can I find the arc length of the polar curve: $r(t)=1+\cos(t)$ using the hint: $\sqrt{1-\cos(t)}\cdot \sqrt{1+\cos(t)} =\sin(t)$ for $0\leq t\leq \pi$ ?

I think the main thing I'm wondering is the factorization, since I'm pretty sure I can use the the formula: $$L=\int_0^{\pi} \sqrt{(dr/dt)^2 +r^2} dt$$ To find the arc length of the upper half of the cardioid and then just multiply it by 2?

So I'm not sure how I can use the hint when I got $$\int_0^{\pi}\sqrt{(-\sin(t))^2 +(1+\cos(t))^2} dt.$$

• As regards the hint, see my P.S. – Robert Z Sep 3 '16 at 19:03

The length $L$ of the cardioid is given by \begin{align*}L&=2\int_0^{\pi} \sqrt{(dr/dt)^2 +r^2} dt\\ &= 2\int_0^{\pi}\sqrt{(-\sin(t))^2 +(1+\cos(t))^2} dt \\&=2\int_0^{\pi}\sqrt{\sin^2(t) +1+2\cos(t)+\cos^2(t)} dt \\&=2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt =4\int_0^{\pi}\cos(t/2) dt=8[\sin(t/2)]_0^{\pi}=8. \end{align*}
P.S. As regards the hint, note that $$2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt = 2\sqrt{2}\int_0^{\pi}\frac{\sin(t)}{\sqrt{1-\cos(t)}} dt= 4\sqrt{2}\int_0^{\pi}\frac{d(1-\cos(t))}{2\sqrt{1-\cos(t)}}\\ =4\sqrt{2}[\sqrt{1-\cos(t)}]_0^{\pi}=8.$$