Problem source

By means of the change of variable $\theta=\frac{1}{4}\pi-\phi,$
or otherwise, show that 
\[
\int_{0}^{\frac{1}{4}\pi}\ln(1+\tan\theta)\,\mathrm{d}\theta=\tfrac{1}{8}\pi\ln2.
\]
Evaluate 
\[
{\displaystyle \int_{0}^{1}\frac{\ln(1+x)}{1+x^{2}}\,\mathrm{d}x}\qquad\mbox{ and }\qquad{\displaystyle \int_{0}^{\frac{1}{2}\pi}\ln\left(\frac{1+\sin x}{1+\cos x}\right)\,\mathrm{d}x}.
\]

Solution source

\begin{align*}
&& I  &= \int_0^{\frac14\pi} \ln (1 + \tan \theta) \d \theta \\
\theta = \tfrac14\pi - \phi, \d \theta = -\d\phi: &&&= \int_0^{\frac14 \pi} \ln ( 1 + \tan (\tfrac14\pi - \phi)) \d \phi \\
&&&= \int_0^{\frac14 \pi} \ln \left ( 1 + \frac{1 - \tan \phi}{1+\tan \phi} \right) \d \phi \\
&&&= \int_0^{\frac14 \pi} \ln \left ( \frac{2}{1+\tan \phi} \right) \d \phi \\
&&&= \tfrac14 \pi \ln 2 - I \\
\Rightarrow && I &= \tfrac18\pi \ln 2
\end{align*}

\begin{align*}
&& J &= \int_0^1 \frac{\ln(1+x)}{1+x^2} \d x \\
x= \tan \theta \d \theta, \d \theta = \frac{\d x}{1+x^2} &&&= \int_0^{\frac14 \pi} \ln(1 + \tan \theta) \d \theta \\
&&&= \tfrac18 \pi \ln 2 
\end{align*}

\begin{align*}
&& K &= \int_0^{\frac12 \pi} \ln \left ( \frac{1 + \sin x}{1 + \cos x} \right) \d x \\
y = \tfrac12\pi - x, \d y = -\d x: &&&= \int_0^{\frac12\pi} \ln \left ( \frac{1+\cos y}{1+\sin y}\right) \d y \\
&&&= -K \\
\Rightarrow && K &= 0

\end{align*}