Evaluate: $\int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ Evaluate: $ \displaystyle \int_0^{\pi} \ln \left( \sin \theta \right) d\theta$ using Gauss Mean Value theorem.
Given hint: consider $f(z) = \ln ( 1 +z)$.
EDIT:: I know how to evaluate it, but I am looking if I can evaluate it using Gauss MVT.
ADDED:: Here is what I have got so far!!
$$\ln 2 = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{i \theta}) d\theta = \frac{1}{2 \pi } \int_0^{2\pi } \log(2+e^{-i \theta}) d\theta$$
Hence, $ \displaystyle 2 \ln 2 = \frac{1}{2 \pi } \int_{0}^{2 \pi} \log(5 + 4 \cos \theta )d \theta  = \frac{1}{\pi} \int_0^{\pi} \log(1 + 8 \cos^2 \theta) d \theta$,  now to problem is how to reduce it to the above form?
 A: $$I=\displaystyle \int_0^{\pi} \ln \left( \sin \theta \right)\cdot  d\theta$$
$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \right) \cdot d\theta$$
$$I=2\times \displaystyle \int_0^{\pi/2} \ln \left( \cos \theta \right) \cdot d\theta$$
Adding both.
$$I=\displaystyle \int_0^{\pi/2} \ln \left( \sin \theta \times \cos \theta\right) \cdot 
d\theta$$
$$I= \displaystyle \int_0^{\pi/2} \ln \left(2 \sin \theta\times \cos \theta \right) -\ln2 \cdot d\theta$$
$$I=\int_0^{\pi/2}\ln(\sin{2\theta})-\ln2 \cdot d\theta$$
$$\int_0^{\pi/2}\ln(\sin{2\theta})\cdot d\theta=I/2$$
So,
$$I=-2\int_0^{\pi/2}\ln2\cdot d\theta$$
$$I=-{\pi\ln2}$$
A: I got this as the first part of this answer:

Start with
  $$
\begin{align}
\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x
&=\frac12\int_0^\pi\log(\sin(x))\,\mathrm{d}x\\
&=\int_0^{\pi/2}\log(\sin(2x))\,\mathrm{d}x\\
&=\int_0^{\pi/2}\Big(\log(2)+\log(\sin(x))+\log(\cos(x))\Big)\,\mathrm{d}x\\
&=\frac\pi2\log(2)+2\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x\tag{1}
\end{align}
$$
  Therefore,
  $$
\int_0^{\pi/2}\log(\sin(x))\,\mathrm{d}x=-\frac\pi2\log(2)\tag{2}
$$

Thus,
$$
\int_0^\pi\log(\sin(x))\,\mathrm{d}x=-\pi\log(2)
$$

Using Gauss Mean Value
$\mathrm{Re}(\log(z))=\log(|z|)=\log\left(\sqrt{2-2\cos(x)}\right)$
$\hspace{4.5cm}$
$$
\begin{align}
\int_0^\pi\log(\sin(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(\color{#C00000}{\frac12}\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x\\
&=\pi\color{#00A000}{\frac1{2\pi}\int_0^{2\pi}\log\left(\sqrt{2-2\cos(x)}\right)\,\mathrm{d}x}\color{#C00000}{-\pi\log(2)}\\[6pt]
&=\pi\color{#00A000}{\log(1)}-\pi\log(2)\\[12pt]
&=-\pi\log(2)
\end{align}
$$
A: My try:
\begin{align}
\int_0^{\pi}\ln\sin\theta\,d\theta=\frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(-\ln 4+\ln4\cos^2\theta\right)d\theta=-\pi\ln 2+\frac12\int_{-\pi/2}^{\pi/2}\ln(2+2\cos2\theta)d\theta=\\=
-\pi\ln 2+\frac14\underbrace{\int_{-\pi}^{\pi}\left[\ln(1+e^{i\theta})+\ln(1+e^{-i\theta})\right]d\theta}_{=0\;\mathrm{by\; MVT}}=-\pi\ln 2.
\end{align}
A: Here is a solution I wrote for a complex analysis assignment several years ago, I hope it helps.  Basically, we are using the mean value theorem you mention above on a slightly different function, and then separating things to obtain the desired integral.  We have to be careful because we can't exactly integrate $\log(1-u)$ on the circle of radius $1$.

Consider 
  $$
\int_{C_{1-\epsilon}}\frac{\log(1-u)}{u}du
$$
   where $C_{1-\epsilon}$ is the circle of radius $1-\epsilon$. Then
  since $\frac{\log(1-u)}{u}$ is an analytic function in $D_{1-\epsilon}$
  (It has a removable singularity at $u=0$ by the removable singularity
  theorem mentioned last assignment), we see that this contour integral
  will be zero for every $\epsilon>0$. But then notice 
  $$
\int_{C_{1-\epsilon}}\frac{\log(1-u)}{u}du=2i\int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz
$$
  so that 
  $$
\int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz=0
$$
   for every $\epsilon>0$. Since 
  $$
|\int_{0}^{\pi}\log(1-e^{i2z})dz|\leq\int_{0}^{\pi}|\log z|dz+\int_{0}^{\pi}|\log(\pi-z)|dz+\int_{0}^{\pi}|\log\left(\frac{1-e^{i2z}}{z(z-\pi)}\right)|dz
$$
  As $\frac{1-e^{i2z}}{z(z-\pi)}$ has no zeros on $[0,\pi]$ we see
  that it must be bounded below by some constant $c$. Then as it also
  has nontrivial imaginary part on $(0,\pi)$ we see that $\int_{0}^{\pi}|\log\left(\frac{1-e^{i2z}}{z(z-\pi)}\right)|dz<\infty$.
  Then since $\int_{0}^{1}\log xdx=x\log x-x\biggr|_{x=0}^{x=1}=-1<\infty$
  it follows that $\int_{0}^{\pi}|\log z|dz<\infty$ and $\int_{0}^{\pi}|\log(\pi-z)|dz<\infty$
  so that $|\int_{0}^{\pi}\log(1-e^{i2z})dz|<\infty$. Recall $\log$
  is uniformly continuous on any compact set not containing the origin,
  so we can bound the middle of all of these integrals by the same constant.
  Since around $0$ and around $\pi$ the norm of $\log(1-e^{i2z})$
  goes to infinity, we can choose small enough neighborhoods so that
  the norm of $\log(1-(1-\epsilon)e^{i2z})dz$ is bounded above by $|\log(1-e^{i2z})|$
  in these neighborhoods for every $\epsilon>0$. Then applying the
  dominated convergence theorem tells us that 
  $$
\lim_{\epsilon\rightarrow0}\int_{0}^{\pi}\log(1-(1-\epsilon)e^{i2z})dz=\int_{0}^{\pi}\log(1-e^{i2z})dz=0.
$$
  Now we have the identity
  $$
1-e^{-2iz}=-2ie^{iz}\sin z
$$
  so that
  $$
0=\int_{0}^{\pi}\log(\sin z))dz+\int_{0}^{\pi}\log(e^{iz})dz+\int_{0}^{\pi}\log(-2i)dz.
$$
  By choosing the principal branch of the logarithm we then have 
  $$
\int_{0}^{\pi}\log(\sin z))dz=-\left(\int_{0}^{\pi}izdz+\int_{0}^{\pi}-\frac{\pi i}{2}dz+\int_{0}^{\pi}\log(2)dz\right)
$$
  $$
=-\left(\frac{i\pi^{2}}{2}+-\frac{\pi^{2}i}{2}dz+\pi\log(2)dz\right)=-\pi\log2.
$$ By substituting $z=\pi x$ we see that $\int_{0}^{\pi}\log(\sin z))dz=\pi\int_{0}^{1}\log(\sin\pi x))dx$
  so that we are able to conclude 
  $$
\int_{0}^{1}\log(\sin\pi x))dx=-\log2
$$
  as desired.

A: This is just for fun. 
It is well known that  $\Pi_{1\le k<n}\sin \frac{k\pi}{n}=\frac{n}{2^{n-1}}$. So $\int_0^\pi \ln \sin x dx =\lim_{n\to \infty}\frac{\pi}{n}\ln \Pi_{1\le k<n} \sin \frac{k\pi}{n}=-\pi\ln 2 $.
