How can the arc length of a cycloid be calculated using parametric equations?

[John O' Meara]

Integrate $$\int_C \sec^2 z dz \ \mbox{any path from } \ \frac{\pi}{4} \mbox{ to } \frac{\pi\iota}{4} \\ \sec^2 z = \frac{1}{\cos^2 z} \ \mbox{ which is equal to } \ \frac{1}{2(1+\cos 2z)} \\$$ Therefore $$\frac{2}{1+\cos 2z} = \frac{2}{1 + \cos2x\cosh2y -\iota \sin2x\sinh2y}\\$$
How do you split this fraction up to differentiate it w.r.t., $$u_x \mbox{,} v_y \\$$ i.e., the Cauchy - Riemann equations? It seems to me that $$sec^2 z$$ is not analytic.
Therefore integrate as follows:$$z(t) = t + (\frac{\pi}{4} -t)\iota \ \frac{dz}{dt} = 1 - \iota \\$$ Therefore the integral is $$\int_0^{\frac{\pi}{4}} \sec^2(t+(\frac{\pi}{4}-t)\iota)(1 - \iota ) dt\\$$. Am I correct so far? Is z substituted in $$\sec^2z \$$ done correctly. Thanks for helping.

 

[Dick]

If f(z) is analytic, then f(z)^2 is analytic and 1/f(z) is analytic (away from poles). Since cos(z) is analytic, there is no point in checking CR for sec(z)^2. They will be satisfied.

 

[John O' Meara]

Cycloid question

The parametric equations of a cycloid are $$x=a(\theta - \sin\theta) \mbox{ and } y=a(1-\cos\theta) \\$$ where a is a constant. Show that S^2=8ay, where s is the arc length measured from the point theta = 0.

$$\frac{dx}{d\theta} = a(1 - \cos\theta) \ \frac{dy}{d\theta} = a\sin\theta \\$$. Then $$\int \sqrt{a^2(1-\cos\theta)^2 + a^2\sin^2 \theta} d\theta \\$$. Now this gives $$\int \sqrt{2}a\sqrt{1 - \cos\theta} d\theta\\$$. I could do with a hint as to, how to solve this integral. Can I simply say $$s^2=2a^2 \int (1 - \cos\theta)d\theta \\$$. I certainly think not. Thanks for helping.