Integrating Square Root of Rational Trigonometric Equation _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.

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:

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