5

Problem

Show that
$$\int_k^\pi \sqrt{\frac{1-\cos x}{\cos k-\cos x}} \, dx = \pi$$
for all $0\leq k<\pi$.

Remark
I was trying to prove the isochronous property of the cycloid curve and I ended up with this integral. I don't know how to start. I'm pretty confident that the answer is $\pi$ because I tried plugging in some values of $k$ in Maple and it always returned approximately $\pi$.

Need help, please.

Michael Hardy
  • 1
Nur Nordin
  • 161
  • 1
    Show the derivative of both sides with respect to $k$ is $0$... – Simply Beautiful Art May 27 '17 at 16:32
  • $$\begin{align} \frac d {dk} \int_{a(k)}^{b(k)} f(k,x) , dx & = f(k,b(k)) \cdot \frac{d}{dk} b(k) - f(k,a(k)) \cdot \frac d {dk} a(k) + \int_{a(k)}^{b(k)} \frac \partial {\partial k} f(k,x), dx \ \ & = f(k,\pi) \cdot \frac d {dk}\pi - f(k,k)\cdot \frac d {dk} k + \int_k^\pi \frac \partial {\partial k} f(k,x) , dx \ \ & = \sqrt{\frac{1-\cos\pi}{\cos k - \cos\pi}} \cdot \frac d {dk} \pi - \underbrace{\sqrt{\frac{1-\cos k}{\cos k - \cos k}} \cdot \frac d {dk} k}_{\text{questionable!}} + \int_k^\pi \frac \partial {\partial k} \sqrt{\frac{1-\cos x}{\cos k - \cos x}} , dx \end{align}$$ – Michael Hardy May 27 '17 at 17:12
  • My comment above should not be taken to mean one cannot find this derivative, but that it will take more work than just mechanically applying this formula. The first term in the last line is obviously zero. Might it be that the second and third terms cancel out in the limit as the thing plugged in in place of $x$ approaches $k\text{?} \qquad$ – Michael Hardy May 27 '17 at 17:15

1 Answers1

4

Well, if you substitute:

$$\text{u}:=\frac{\sqrt{2}\cdot\cos\left(\frac{x}{2}\right)}{\sqrt{1+\cos\left(\text{k}\right)}}\tag1$$

You end up with:

$$\mathscr{I}_{\space\text{k}}:=\int_\text{k}^\pi\sqrt{\frac{1-\cos\left(x\right)}{\cos\left(\text{k}\right)-\cos\left(x\right)}}\space\text{d}x=-2\int_{\frac{\sqrt{2}\cdot\cos\left(\frac{\text{k}}{2}\right)}{\sqrt{1+\cos\left(\text{k}\right)}}}^0\frac{1}{\sqrt{1-\text{u}^2}}\space\text{d}\text{u}=$$ $$-2\cdot\left\{\arcsin\left(0\right)-\arcsin\left(\frac{\sqrt{2}\cdot\cos\left(\frac{\text{k}}{2}\right)}{\sqrt{1+\cos\left(\text{k}\right)}}\right)\right\}=2\cdot\arcsin\left(\frac{\sqrt{2}\cdot\cos\left(\frac{\text{k}}{2}\right)}{\sqrt{1+\cos\left(\text{k}\right)}}\right)=$$ $$\pi\cdot\sqrt{\cos^2\left(\frac{\text{k}}{2}\right)}\cdot\sec\left(\frac{\text{k}}{2}\right)\tag2$$

And, use:

$$\cos\left(\frac{\text{k}}{2}\right)\cdot\sec\left(\frac{\text{k}}{2}\right)=1\tag3$$

Jan Eerland
  • 29,457