Can someone give me a hint on how to evaluate the closed form of the following integral? According to Wolfram Alpha, this evaluates to $\dfrac{\pi}8\log(2)$.
$$\int\limits_0^{\pi/4}\log(1+\tan x)\,\mathrm dx$$
Thanks.
p.s - Hints are preferred over complete solutions.
$$I=\int^{\frac{\pi }{4} }_{0}\ln\left( 1+\tan x \right)\ dx=\int ^{\frac{\pi }{4} }_{0}\ln\left( 1+\tan \left( \frac{\pi }{4} -x\right) \right)\ dx=\int^{\frac{\pi }{4} }_{0}\ln\left( 2\right)\ dx-I$$ So we have $$2I=\ln\left( 2\right) \int ^{\frac{\pi }{4} }_{0}\ dx=\ln\left( 2\right) \times \frac{\pi }{4} $$ Hence we get$$ I= \boxed{\frac{\pi }{8} \ln\left( 2\right)} $$
Hint. By the change of variable $$ x=\frac{\pi}4-u, \qquad dx=-du, \qquad 1+ \tan x=? \qquad \log(1+ \tan x)=? $$
The post asks for a hint, but that was quite a while ago. For the sake of the site, I think a full answer might be useful:
$$
\begin{align}
&\int_0^{\pi/4}\log(1+\tan(x))\,\mathrm{d}x\\
&=\int_0^{\pi/4}\log(\cos(x)+\sin(x))\,\mathrm{d}x-\int_0^{\pi/4}\log(\cos(x))\,\mathrm{d}x\tag1\\
&=\int_0^{\pi/4}\log(\sqrt2\cos(\pi/4-x))\,\mathrm{d}x-\int_0^{\pi/4}\log(\cos(x))\,\mathrm{d}x\tag2\\
&=\frac\pi8\log(2)+\int_0^{\pi/4}\log(\cos(x))\,\mathrm{d}x-\int_0^{\pi/4}\log(\cos(x))\,\mathrm{d}x\tag3\\[6pt]
&=\frac\pi8\log(2)\tag4
\end{align}
$$
Explanation:
$(1)$: $1+\tan(x)=\frac{\cos(x)+\sin(x)}{\cos(x)}$
$(2)$: $\cos(x)+\sin(x)=\sqrt2\cos(\pi4-x)$
$(3)$: integrate the constant $\log(\sqrt2)$
$\phantom{\text{(3):}}$ substitute $x\mapsto\pi/4-x$ in the left integral
$(4)$: cancel
Note
\begin{align}&\int^{\frac{\pi }{4} }_{0}\ln( 1+\tan x )\ dx =\frac12 \int^{\frac{\pi }{4} }_{0}\ln( 1+\tan x )^2\ dx\\ =& \frac12 \int^{\frac{\pi }{4} }_{0}\ln\sec^2x\ dx + \frac12 \int^{\frac{\pi }{4} }_{0}\underset{2x\to \frac\pi2-2x}{\ln(1+\sin2x)\ dx}\\ = &\frac12 \int^{\frac{\pi }{4} }_{0}\ln(\sec^2x(1+\cos2x))\ dx =\frac12 \int^{\frac{\pi }{4} }_{0}\ln2\ dx=\frac\pi8\ln2 \end{align}
