Integral $\int _{0}^{\pi /2}x\cos (8x)\ln \tan x \, \text{d}x$ Inspired by this topic, can I easily prove the result below?
$$\int _{0}^{\pi /2}x\cos (8x)\ln \tan x \, \text{d}x=\frac{13}{36}$$
The elementary antiderivative exists, but it seems masochist if one wants compute it.
 A: As you wrote, a CAS can find the antiderivative $$f(x)=\int x \cos(nx)\log (\tan (x))\,dx$$ and it is quite ugly (involving hypergeometric series with arguments $e^{4ix}$).
So, it is clear that, if $n$ is a multiple of $4$ , there will be a lot of simplifications for $$g(n)=\int_0^{\frac\pi 2} x \cos(nx)\log (\tan (x))\,dx$$ as shown below for the first ones.
$$\left(
\begin{array}{cc}
 n & g(n) \\
 4 & \frac{3}{4} \\
 8 & \frac{13}{36} \\
 12 & \frac{211}{900} \\
 16 & \frac{7613}{44100} \\
 20 & \frac{270407}{1984500}
\end{array}
\right)$$ What is also interesting is that for other even values of $n$, the result is just a multiple of ${\pi^2}$
$$\left(
\begin{array}{cc}
 n & g(n) \\
 2 & -\frac{\pi ^2}{8} \\
 6 & -\frac{\pi ^2}{24} \\
 10 & -\frac{\pi ^2}{40} \\
 14 & -\frac{\pi ^2}{56} \\
 18 & -\frac{\pi ^2}{72}
\end{array}
\right)$$ the denominator being just $4n$.
For sure, for these specific cases and in particular the one you posted, as @tired commented, integration by parts leads to "quite simple" result.
A: $\displaystyle \int x \cos(8x) \ln(\tan x) dx =$ $\displaystyle \frac{1}{576}(9\ln(\cos x)+72x\sin(8x)\ln(\tan x)+9\cos(8x)\ln(\tan x)-$ $\displaystyle 9\ln(\sin x)-144x\sin(2x)-48x\sin(6x)-90\cos(2x)-14\cos(6x)) + C$
Because of $\enspace\displaystyle \lim_{x\to\pi/2} (\cos x) (\tan x)^{8x \sin(8x) + \cos (8x)} = 1 \enspace$ and $\enspace\displaystyle \lim_{x\to 0} \frac{(\tan x)^{\cos (8x)}}{\sin x} = 1$ 
we get $\enspace\displaystyle \int\limits_0^{\pi/2} x \cos(8x) \ln(\tan x) dx =\frac{2}{576}(90+14)=\frac{13}{36}$ .
