# $\int_0^{2\pi} \sqrt{1-\cos(x)}\,dx = 4\sqrt{2}$. Why?

According to the textbook, and Wolfram Alpha the above is correct.

Here is the step by step procedure from Wolfram Alpha for evaluating the indefinite integral:

Take the integral: $$\int\sqrt{1-\cos(x)}\,dx$$ For the integrand $\sqrt{1-\cos(x)}$, substitute $u=1-\cos(x)$ and $du=\sin(x)\,dx$: $$=\int-\frac{1}{\sqrt{2-u}}\,du$$ Factor out constants: $$=-\int\frac{1}{\sqrt{2-u}}\,du$$ For the integrand $1/\sqrt{2-u}$, substitute $s=2-u$ and $ds=-du$: $$=\int\frac{1}{\sqrt{s}}\,ds$$ The integral of $1/\sqrt{s}$ is $2\sqrt{s}$: $$=2\sqrt{s}+\text{constant}$$ Substitute back for $s=2-u$: $$=2\sqrt{2-u}+\text{constant}$$ Substitute back for $u=1-\cos(x)$: $$=2\sqrt{\cos(x)+1}+\text{constant}$$ Which is equivalent for restricted $x$ values to: $$\boxed{=-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}}$$

I understand up to the below (which is a valid solution to the integral): $$2\sqrt{\cos(x)+1}+\text{constant}$$

However, if you evaluate this at $2\pi$ and $0$, you get the same thing, so the definite integral evaluates to zero.

After, you transform the above to: $$-2\sqrt{1-\cos(x)}\cot\big(\frac{x}{2}\big)+\text{constant}$$

The expression is indeterminate at $2\pi$ and $0$ of the form $0 \times \infty$. So I guess you would set up a limit and then use L'Hospital's rule to evaluate the expression at $2\pi$ and $0$ and get the answer to the definite integral?

In any case, all this seems strange. Why should the definite integral evaluated one way give $0$, and in another way give something else?

• You may want to find the anti derivative by multiplying top and bottom by the squareroot of the conjugate. You will end up with the absolute value of the sine term on top and the square root of 1+cosx in the bottom. This integrates easily. Mind the absolute value though as you apply your upper and lower limit. Commented Oct 30, 2013 at 19:57
• Ahh ok that makes sense. So to get rid of the absolute value sign, you can split up the integral from $0$ to $\pi$ and $\pi$ to $2\pi$ Commented Oct 30, 2013 at 20:02
• Yes, you ought to split that integral. Commented Oct 30, 2013 at 20:05

$$\int_0^{2\pi}\sqrt{1-\cos x}\,dx=\int_0^{2\pi}\sqrt{2\sin^2\frac{x}{2}}\,dx =\sqrt{2}\int_0^{2\pi}\sin\frac{x}{2}\,dx=-2\sqrt{2}\cos\frac{x}{2}\Big|_0^{2\pi}=4\sqrt{2}.$$
• @RonGordon $x/2 \in [0,\pi] \Rightarrow \sin(x/2) \ge 0$. Commented Oct 30, 2013 at 20:49
• Good old $\cos(2x) = \cos^2(x)-\sin^2(x)= 1-2\sin^2(x) = 2\cos^2(x)-1$. Commented Oct 31, 2013 at 1:25
Your $u$-substitutions should be injective on their interval of evaluation. Otherwise, you risk running into exactly this sort of issue.
Note that \begin{align}|\sin x| &= \sqrt{\sin^2 x}\\ &= \sqrt{1-\cos^2 x}\\ &= \sqrt{1-\cos x}\sqrt{1+\cos x}\\ &=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)},\end{align} so if you want to use $u=1-\cos x$, then $$\frac{du}{dx}=\sin x=\begin{cases}|\sin x|=\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & 0\le x\le \pi\\-|\sin x|=-\sqrt{1-\cos x}\sqrt{2-(1-\cos x)} & \pi\le x\le2\pi,\end{cases}$$ so \begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx+\int_\pi^{2\pi}\frac{|\sin x|}{\sqrt{2-(1-\cos x)}}\,dx\\ &= \int_0^\pi\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}-\int_\pi^{2\pi}\frac{\sin x\,dx}{\sqrt{2-(1-\cos x)}}\\ &= \int_0^2\frac{du}{\sqrt{2-u}}-\int_2^0\frac{du}{\sqrt{2-u}}\\ &= 2\int_0^2\frac{du}{\sqrt{2-u}}.\end{align} At that point, we can use that antiderivative, with no need to resubstitute.
Alternately, you could note that $\cos(2\pi-x)=\cos x$, so \begin{align}\int_0^{2\pi}\sqrt{1-\cos x}\,dx &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos x}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx+\int_\pi^{2\pi}\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_{2\pi}^\pi\sqrt{1-\cos(2\pi-x)}\,dx\\ &= \int_0^\pi\sqrt{1-\cos x}\,dx-\int_0^\pi\sqrt{1-\cos x}\frac{d(2\pi-x)}{dx}\,dx\\ &= 2\int_0^\pi\sqrt{1-\cos x}\,dx,\end{align} at which point you can use your $u$-substitution without fear, since the cosine function is injective on $[0,\pi]$.