Arc Length parametric curve I have the following curve:
$$x = \cos(t)$$
$$y = t - \sin(t)$$
$$0 \leq t \leq 2\pi$$
I have to draw the graph, point the direction and find its length. The solved the first two questions. The problem emerges when I set up the integral for the arc length.
$$\frac{\partial x}{\partial t} = -\sin(t)$$
$$\frac{\partial y}{\partial t} = 1 - \cos(t)$$
$$\int ^{2\pi}_0{\sqrt{(-\sin^2t)+(1-\cos(t))^2}dt}$$
After simplification:
$$\int ^{2\pi}_0{\sqrt{2-2\cdot \cos(t)}dt}$$
But how to proceed further?
 A: Hint: $$\begin{align}\int_0^{2\pi} \sqrt {2 (1 - \cos t)} dt &= \int_0^{2\pi} \sqrt2 \sqrt{2 \sin^2 \left(\frac{t}{2} \right)} dt \\&= 2 \int_0^{2\pi}\sin \left( \frac{t}{2}\right) dt \\&= \ldots\end{align} $$
A: Split the limits into four intervals i.e $ [0,\pi/2], [\pi/2,\pi], [\pi,3\pi/2] and [3\pi/2,2\pi]$.
Let $A$ = $\int ^{\pi/2}_0{\sqrt{2-2\cdot \cos(t)}dt}$ be the arc length in domain $[0,\pi/2]$.
Let $B$ = $\int ^{\pi}_{\pi/2}{\sqrt{2-2\cdot \cos(t)}dt}$ be the arc length in domain $[\pi/2,\pi]$.
Let $C$ = $\int ^{3\pi/2}_{\pi}{\sqrt{2-2\cdot \cos(t)}dt}$ be the arc length in domain $[\pi,3\pi/2]$.
Let $D$ = $\int ^{2\pi}_{3\pi/2}{\sqrt{2-2\cdot \cos(t)}dt}$ be the arc length in domain $[3\pi/2,2\pi]$.
Let $I$ = $A+B+C+D$ is the solution which is required to be calculated
Since $cos(t)$ is symmetric about the line $t-\pi=0$.This implies $\sqrt{2-2\cdot \cos(t)}$  is also symmetric about the line $t-\pi=0$.Thus $A+B=C+D$
This implies $I=2*(A+B)$
$I=2*(\int ^{\pi/2}_0{\sqrt{2-2\cdot \cos(t)}dt}+\int ^{\pi}_{\pi/2}{\sqrt{2-2\cdot \cos(t)}dt})$
$I=2*\int ^{\pi}_0{\sqrt{2-2\cdot \cos(t)}dt}$
Let $x=cos(t)$
Then $dx=-sin(t) dt$
As $sin(t)=\sqrt{1-\cos^2(t)}$
Thus $sin(t)=\sqrt{1-x^2}$
Putting the value of $cos(t)$ and $dt$ in $I$ and changing the limits we get, 
$I=-2*\int ^{-1}_{1}{\sqrt{({2-2x})/({1-x^2})}dx}$
$I=2\sqrt{2}*\int ^{1}_{-1}{\sqrt{({1-x})/({1-x^2})}dx}$
$I=2\sqrt{2}*\int ^{1}_{-1}{\sqrt{1/({1+x})}dx}$
Integrating it and putting limits we get,
$I=8$ as the required answer.
