Problem source

\begin{questionparts}
\item Prove that 
\[
\int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x=\int_{0}^{\frac{1}{2}\pi}\ln(\cos x)\,\mathrm{d}x=\tfrac{1}{2}\int_{0}^{\frac{1}{2}\pi}\ln(\sin2x)\,\mathrm{d}x-\tfrac{1}{4}\pi\ln2
\]
and 
\[
\int_{0}^{\frac{1}{2}\pi}\ln(\sin2x)\,\mathrm{d}x=\tfrac{1}{2}\int_{0}^{\pi}\ln(\sin x)\,\mathrm{d}x=\int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x.
\]
Hence, or otherwise, evaluate ${\displaystyle \int_{0}^{\frac{1}{2}\pi}\ln(\sin x)\,\mathrm{d}x.}$
\textit{You may assume that all the integrals converge.}
\item Given that $\ln u< u$ for $u\geqslant1$ deduce that 
\[
\tfrac{1}{2}\ln x < \sqrt{x}\qquad\mbox{ for }\quad x\geqslant1.
\]
Deduce that $\dfrac{\ln x}{x}\rightarrow0$ as $x\rightarrow\infty$ and that $x\ln x\rightarrow0$ as $x\rightarrow0$ through positive values. 
\item Using the results of parts \textbf{(i) }and \textbf{(ii)}, or otherwise, evaluate ${\displaystyle \int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x.}$
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
u = \frac{\pi}{2} - x :&& \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \int_{\frac12\pi}^0 \ln (\cos u) (- 1)\d u  \\
&&&= \int_0^{\frac12 \pi} \ln (\cos x) \d x \\
\Rightarrow && 2 \int_0^{\tfrac12 \pi} \ln (\sin x) \d x  &= \int_0^{\tfrac12 \pi} \ln (\sin x) \d x +\int_0^{\tfrac12 \pi} \ln (\cos x) \d x \\
&&&=  \int_0^{\tfrac12 \pi}\left (\ln (\sin x)+  \ln (\cos x) \right) \d x \\
&&&= \int_0^{\frac12 \pi} \ln \left (\frac12 \sin 2x \right) \d x \\
&&&= \int_0^{\frac12 \pi} \left ( \ln \left (\sin 2x \right) - \ln 2 \right)\d x \\
&&&=  \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x  - \frac{\pi}{2} \ln 2\\
\Rightarrow &&  \int_0^{\tfrac12 \pi} \ln (\sin x) \d x &= \frac12 \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x  - \frac{\pi}{4} \ln 2
\end{align*}

\begin{align*}
u = 2x, \d u = 2 \d x &&  \int_0^{\frac12 \pi} \ln \left (\sin 2x \right)\d x &= \int_0^{\pi} \ln (\sin u) \frac12 \d u \\
&&&= \frac12 \int_0^{\pi} \ln (\sin u) \d u \\
&&&=\frac12 \left ( \int_0^{\pi/2} \ln (\sin u) \d u + \int_{\pi/2}^{\pi} \ln (\sin u) \d u \right)\\
&&&= \int_0^{\pi/2} \ln (\sin u) \d u \\
\Rightarrow && I &= \frac12 I - \frac14 \pi \ln 2 \\
\Rightarrow && I &= -\frac12 \pi \ln 2 
\end{align*}

\item \begin{align*}
&& \ln u &< u & \quad (u \geq 1)\\
\underbrace{\Rightarrow}_{u = \sqrt{x}} && \ln \sqrt{x} &< \sqrt{x} \\
\Rightarrow && \frac12 \ln x &< \sqrt{x} \\
\Rightarrow && \frac{\ln x}{x} &< \frac{2\sqrt{x}}{x} \\
&&&= \frac{2}{\sqrt{x}} \\
&&&\to 0 & (x \to \infty)
\\
&& x \ln x &= \frac{\ln 1/y}{y} \\
&&&= -\frac{\ln y}{y} \\
&&&\to 0 & (y \to \infty, x \to 0)
\end{align*}

\item \begin{align*}
\int_{0}^{\frac{1}{2}\pi}x\cot x\,\mathrm{d}x &= \left [ x \ln(\sin x) \right]_0^{\pi/2} - \int_0^{\pi/2} \ln (\sin x) \d x \\
&= \left ( \frac{\pi}{2} \ln 1 - \lim_{x \to 0} x \ln (\sin x)  \right) - \left ( -\frac12 \pi \ln 2  \right) \\
&= \frac12 \pi \ln 2

\end{align*}

\end{questionparts}