In my post,
I had found an elegant integral
$$\int_{0}^{\frac{\pi}{4}} \ln (\tan x)~ d x=-G. $$
I then try to generalise the integral
$$
I_{n}=\int_{0}^{\frac{\pi}{4}} \ln ^{n}(\tan \theta) d \theta, ~~~~\textrm{ where }n\in N,
$$
by letting $e^{-x}=\tan \theta$. Consequently $-e^{-x} d x=\left(1+e^{-2 x}\right) d \theta$ converts the integral to
$$
\begin{aligned}
I_{n} &=\int_{\infty}^{0} \frac{\ln ^{n}\left(e^{-x}\right)\left(-e^{-x}\right)}{1+e^{-2 x}} d x =\int_{0}^{\infty} \frac{(-x)^{n} e^{-x}}{1+e^{-2 x}} d x
\end{aligned}
$$
Expanding the integrand into a power series yields
$$
\begin{aligned}
I_{n} &=(-1)^{n} \int_{0}^{\infty} \sum_{k=0}^{\infty}(-1)^{k} x^{n} e^{-(2 k+1) x} d x\\&=(-1)^{n} \sum_{k=0}^{\infty}(-1)^{k} \int_{0}^{\infty} x^{n} e^{-(2 k+1) x} d x
\end{aligned}
$$
Letting $(2k+1)x\mapsto x$ transforms the integral into a Gamma function.
$$
\\ \boxed{\int_{0}^{\frac{\pi}{4}} \ln ^{n}(\tan \theta) d \theta =(-1)^{n} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{\left({2 k+1}\right)^{n+1}} \int_{0}^{\infty}x^ne^{-x} d x=(-1)^{n} \beta(n+1) \Gamma(n+1)}
$$
where $\beta(s)$ is the Dirichlet beta function.
For examples, $$
\begin{array}{l}
\displaystyle I_{1}=\int_{0}^{\frac{\pi}{4}} \ln (\tan \theta)d\theta=-\beta(2) \Gamma(2)=-G \\
\displaystyle I_{2}=\int_{0}^{\frac{\pi}{4}} \ln ^{2}(\tan \theta) d\theta =\beta(3) \Gamma(3)=\frac{\pi^{3}}{32} \cdot 2=\frac{\pi^{3}}{16} \\
\displaystyle I_{10}=\int_{0}^{\frac{\pi}{4}} \ln ^{10}(\tan \theta) d\theta =\beta(11) \Gamma(11)=\frac{50521\pi^{11}}{14863564800}\times 10!= \frac{50521 \pi^{11}}{4096} \approx 3.62878 \times 10^{6}
\end{array}
$$
Looking forwards to getting more alternate solutions and opinions from you!
Best Answer
Let $J(s)=\int_0^{\pi/4} \tan^s(\theta)d\theta$
$$I_n=\frac{d^n}{ds^n}J(s)|_{s=0}$$
Let $t=\tan(\theta)$
$$J(s)=\int_0^1 \frac{t^s}{1+t^2}dt=\sum_{k=0}^\infty (-1)^k \int_0^1 t^{2k+s}dt=\sum_{k=0}^\infty \frac{(-1)^k}{2k+1+s} $$
$$\frac{d^n}{ds^n}J(s)=(-1)^n~n!~\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1+s)^{n+1}}$$
$$I_n=(-1)^n~n!~\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^{n+1}}$$