Solve Integral of ln(sin x) from 0 to pi.

[anemone]

Evaluate the integral \(\displaystyle \int_{0}^{\pi} \ln (\sin x)\,dx\).

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[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. Olok
3. lfdahl
3. chisigma

Solution from MarkFL:

Spoiler

We are given to evaluate:

\(\displaystyle I=\int_{0}^{\pi} \ln\left(\sin(x)\right)\,dx\)

By symmetry, we see that we may write:

\(\displaystyle I=2\int_{0}^{\frac{\pi}{2}} \ln\left(\sin(x)\right)\,dx\)

\(\displaystyle I=2\int_{0}^{\frac{\pi}{2}} \ln\left(\cos(x)\right)\,dx\)

Adding these two, and applying a double-angle identity for sine, and the properties of logs, we obtain:

\(\displaystyle I=\int_{0}^{\frac{\pi}{2}} \ln\left(\sin(2x)\right)-\ln(2)\,dx\)

using the substitution:

\(\displaystyle u=2x\,\therefore\,du=2\,dx\)

we then have:

\(\displaystyle I=\frac{1}{2}\int_{0}^{\pi} \ln\left(\sin(u)\right)-\ln(2)\,du\)

Integrating term by term, we find:

\(\displaystyle I=\frac{1}{2}I-\frac{\pi}{2}\ln(2)\)

Solving for $I$, we get:

\(\displaystyle I=-\pi\ln(2)\)

And so, we may conclude:

\(\displaystyle \int_{0}^{\pi} \ln\left(\sin(x)\right)\,dx=-\pi\ln(2)\)

Solution from lfdahl:

Spoiler

\[ \int_0^\pi \ln(\sin(x))dx = \int_0^\pi \ln(2\cos( \frac{x}{2} )\sin( \frac{x}{2} ))dx
\\\\
=\int_0^\pi \ln(2)dx+\int_0^\pi \ln(\cos( \frac{x}{2} ))dx+\int_0^\pi \ln(\sin( \frac{x}{2} )dx
\\\\=\pi \ln(2)+\int_{\frac{\pi}{2}}^{0}\ln(\cos(\frac{\pi -2u}{2})) d(\pi -2u)+\int_{0}^{\frac{\pi}{2}}\ln(\sin(\frac{2v}{2})d(2v)
\\\\
=\pi \ln(2)+2\int_{0}^{\frac{\pi}{2}}\ln(\sin(u))du+2\int_{0}^{\frac{\pi}{2}}\ln(\sin(v))dv
\\\\
=\pi \ln(2)+\int_{0}^{\pi}\ln(\sin(u))du + \int_{0}^{\pi}\ln(\sin(v))dv
\\\\
=\pi \ln(2)+2\int_{0}^{\pi}\ln(\sin(x))dx\]

Hence

\[ \int_{0}^{\pi}\ln(\sin(x))dx=-\pi \ln(2)\]