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I want to show that

$$\int_0^{\pi} \log(2 - 2 \cos x) = 0$$

However, I cannot do this. I tried splitting the integral into $$\int_0^{\pi/3} \log(2 - 2 \cos x)\,dx + \int_{\pi/3}^{\pi} \log(2 - 2 \cos x) \,dx$$ and showing that the two parts were negatives of one another, which did not work. Wolframalpha does not give very a simple antiderivative. I was wondering if there was a nice way to do this.

Other attempts: using $$\int_0^a f(x) \,dx = \int_0^{a} f(a-x) \,dx$$ and trying to change $\cos(x)$ to $\sin(x$ by some substitution like $u = \pi/2 - x$ and trying to get things to cancel.

Quanto
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calculusstudent123
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    $$2sin^{2}(\frac{x}{2})= [1−cos(x)]$$ – user29418 Jul 23 '19 at 01:20
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    It's improper at $0$. – Randall Jul 23 '19 at 01:26
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    $$\ln\left(2 - 2\cos(x)\right) = \ln\left(2\left(1 - \cos(x)\right)\right) = \ln(2) + \ln\left(1 - \cos(x)\right) $$

    Thus,

    $$I = \ln(2)\pi + \int_0^\pi \ln\left(1 - \cos(x)\right):dx $$

    For the remaining integral use the Weirerstrauss Substitution $t = \tan\left( x/2\right)$

     –  Jul 23 '19 at 01:36

4 Answers4

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First we prove

$$\displaystyle\int_0^{\pi/2} \log \left(\sin x\right)\,dx =-\frac12\pi\log2$$

Proof $\ $ From the substitution $x \to \frac{\pi}{2}-x$ , we get $$ \displaystyle\int_0^{\pi/2} \log( \sin x )\,dx = \displaystyle\int_0^{\pi/2} \log ( \cos x )\,dx $$

Thus \begin{align} 2\displaystyle\int_0^{\pi/2} \log( \sin x )\,dx &= \displaystyle\int_0^{\pi/2} \log (\sin x \cos x )\,dx \\ &= \displaystyle\int_0^{\pi/2} \log (\sin 2x )\,dx-\frac{1}{2}\pi\log 2 \\ &=\frac12\displaystyle\int_0^{\pi} \log (\sin x )\,dx-\frac{1}{2}\pi\log 2 \\ &= \displaystyle\int_0^{\pi/2} \log( \sin x )\,dx -\frac{1}{2}\pi\log 2 \end{align} Then we arrive at the conclusion.

To calculate the original integral, we have \begin{align} \int_0^\pi \log(2-2\cos x)\,dx &= \int_0^\pi \log(4\sin^2 \frac{x}{2})\,dx \\ &= 2\pi \log 2 + 2 \int_0^\pi \log (\sin \frac{x}{2}) \,dx \\ &= 2\pi \log 2 + 4\int_0^{\pi/2} \log \left(\sin x\right)\,dx \\ &= 0 \end{align}

Chiquita
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$$\log(2-2\cos x)=\log(2(1-\cos x))=\log2+\log(2\sin^2 x/2)=2\log2+2\log\sin x/2$$

If we show that $\int_0^\pi\log\sin(x/2)~\mathrm dx=-\pi\log2$, we are done.

With $t\mapsto x/2$, the above claim is,

$$\int_0^{\pi/2}\log\sin t~\mathrm dt=-\frac \pi2\log2$$

This is a pretty well known integral. Can you take it from here?

mars
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$$\int_0^\pi \log \left(4\sin^2\left(\frac{x}{2}\right)\right) dx$$ is equivalent. Use logarithm rules to make it - $$\int_0^\pi \left[ \log 4 + 2 \log\left(\sin\left(\frac{x}{2}\right)\right) \right] dx$$

The rightmost part of the integral has already been solved, which is linked here.

J. W. Tanner
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Zach
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Utilize $(1-a)(1+a)=1-a^2$\begin{align} I=& \int_0^{\pi} \log(2 - 2 \cos x)dx \overset{x\to \pi-x}= \int_0^{\pi} \log(2 + 2 \cos x)dx\\ =&\ \frac12 \int_0^{\pi} \log(4 - 4 \cos^2x)dx = \frac12 \int_0^{\pi} \log(2 - 2 \cos2x)\overset{2x\to x}{dx}\\ =&\ \frac14 \int_0^{2\pi} \log(2 - 2 \cos x)dx=\frac12I=0 \end{align}

Quanto
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