Question about the integral of 1-cos(x).

[333happy]

I am given [MATH]\int\sqrt{1-\cos{x}}.[/MATH] I set [MATH]x=2u[/MATH] so that [MATH]dx=2du[/MATH]. Then I get
[MATH] \begin{align*} &2\int\sqrt{1-\cos{2u}}\,du\\ &=2\sqrt{2}\int\sqrt{\frac{1-\cos{2u}}{2}}\,du\\ &=2\sqrt{2}\int\sqrt{\sin^2u}\,du\\ &=2\sqrt{2}\int\sin{u}\,du\\ &=-2\sqrt{2}cos(x/2). \end{align*} [/MATH]It's the same as the answer at the back of the book. My question is, why do we not have plus or minus square root of sin^2 u??? why is it only the positive square root???

 

[Cubist]

The sqrt operation will return the +ve root. Therefore a mistake creeps into your working because...

[math] \sqrt{\sin^{2}\left(u\right)} = \left|\sin\left(u\right)\right| [/math]
If the absolute value symbol is omitted then it would only hold for restricted values of u (where sin(u)>=0). Does the original question restrict the domain of x?

(The work also misses a constant of integration).

 

[Steven G]

Hmm, where did the 2du come from in the 2nd integral? I see that you stated that dx=2du but since the 1st integral has no dx I can't understand where the 2du comes from in that 2nd integral. Maybe you could explain?

 

[Dr.Peterson]

If you're saying that the book gives exactly the same answer you show, even though you know yours is wrong, my answer is that they're wrong. WolframAlpha gives something different: https://www.wolframalpha.com/input/?i=integrate+sqrt(1−cosx)

I don't always trust them, but this looks right, especially comparing their graph to the graph of your answer.

I suspect that whoever wrote the book's answer failed to consider the domain, even if they used a different method. That's easy to do. Or maybe the problem does specify a restricted domain, and you failed to mention it.

But your method is very creative. As I understand it, you are intentionally aiming at the half-angle formula.

 

[pka]

Please read reply #4 by Prof Peterson, then read this:

 

[Cubist]

The Wolfram step-by-step posted by pka seems quite strange. It follows a method similar to post#1 (making the same error that I pointed out in post#2) and then it suddenly corrects this with an unexplained jump at the last step.

I spotted (by graphing) that Wolfram's final result can be made continuous by adding a floor term...

[math] 4\sqrt{2}\left\lfloor{\frac{x}{2\pi}}\right\rfloor-2\sqrt{1-\cos\left(x\right)}\cdot \cot\left(\frac{x}{2}\right) + constant[/math]

 

[Cubist]

I spotted (by graphing) that Wolfram's final result can be made continuous by adding a floor term...

Actually "continuous" was the wrong word to use. Adding the floor term removes the large steps that appear in the Wolfram result but there remain point discontinuities at n*2*pi (integer n) due to the "cot".