First, we will use complex analysis to evaluate the integral of interest. To that end we write for $n>m$
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac12\oint_C\frac{z^{2m}}{1+z^{2n}}\,dx\\\\
&=\pi i \sum_p\text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right) \tag 1
\end{align}$$
where $C$ is the classical "infinite semi-circular contour" closed in the upper-half plane and $z_p$ are the poles of $\frac{z^{2m}}{1+z^{2n}}$ in the upper-half plane. Note that since $n>m$ the contribution from the integration over the "infinite" semi-circle is zero.
To calculate the poles in the upper-half plane, we simply find the roots of the denominator $1+z^{2n}$ in the upper-half plane, which are $z_p=e^{i(2p+1)\pi/2n}$ for $0\le p\le n-1$.
To calculate the residues, we use L'Hospital's Rule to obtain
$$\begin{align}
\text{Res}\left(\frac{z^{2m}}{1+z^{2n}},z=z_p\right)&=\lim_{z\to z_p}\left(\frac{(z-z_p)z^{2m}}{1+z^{2n}}\right)\\\\
&=\frac{z_p^{2m}}{2nz_p^{2n-1}}\\\\
&=\frac1{2n}z_p^{2m-2n+1}\\\\
&=\frac1{2n}e^{i(2p+1)(2m-2n+1)\pi/2n}\\\\
&=-\frac1{2n}e^{i(2p+1)(2m+1)\pi/2n} \tag 2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=-\frac{\pi i}{2n} \sum_{p=0}^{n-1}e^{i(2p+1)(2m+1)\pi/2n}\\\\
&=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n} \sum_{p=0}^{n-1}\left(e^{i(2m+1)\pi/n}\right)^p\\\\
&=-\frac{\pi i}{2n}e^{i(2m+1)\pi/2n}\frac{2}{1-e^{i(2m+1)\pi/n}}\\\\
&=\frac{\pi}{2n\sin\left(\frac{(2m+1)\pi}{2n}\right)} \tag 3
\end{align}$$
Enforcing the substitution $1+x^{2n}\to 1/x$ yields
$$\begin{align}
\int_0^\infty\frac{x^{2m}}{1+x^{2n}}\,dx&=\frac1{2n} \int_0^1 x^{-(2m+1)/n}\,(1-x)^{((2m+1)/2n)-1}\,dx\\\\
&=\frac1{2n}B\left(\frac{2m+1}{2n},1-\frac{2m+1}{2n}\right)\\\\
&=\frac1{2n}\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right) \tag 4\\\\
\end{align}$$
Comparing $(3)$ and $(4)$ we find that
$$\Gamma\left(\frac{2m+1}{2n}\right)\,\Gamma\left(1-\frac{2m+1}{2n}\right)=\frac{\pi}{\sin\left(\frac{(2m+1)\pi}{2n}\right)} $$
which using the density of the rational numbers and then extending by analytic continuation becomes
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$$
as was to be shown!
They used the transformation $ t = -\log(u)$. Then you should note that
$$ e^{-t}\,\mathrm{d}t = - \,\mathrm{d} u $$
Moreover, the integration bounds change from $(0,\infty) \mapsto (1,0)$. Now, you should reverse the order of integration to get the bounds $(0,1)$. This gives a minus sign which cancels against the one we had above.
I hope this helps. If you have any further questions, feel free to ask them!
Best Answer
Let $u=\cos(x)$, then $$ \begin{align} \int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x &=\frac12\int_{-1}^1\log\left(1-u^2\right)\,\mathrm{d}u\\ &=\frac12\left(\int_{-1}^1\log(1-u)\,\mathrm{d}u+\int_{-1}^1\log(1+u)\,\mathrm{d}u\right)\\ &=\int_0^2\log(v)\,\mathrm{d}v\\[3pt] &=\left.v\log(v)-v\right]_0^2\\[9pt] &=2\log(2)-2\tag{1} \end{align} $$
Using the comment by Jack D'Aurizio $$ \begin{align} &\int_0^1\sin(\pi x)\log(\Gamma(x))\,\mathrm{d}x\tag{2}\\ &=\int_0^1\sin(\pi x)\log(\Gamma(1-x))\,\mathrm{d}x\tag{3}\\ &=\frac12\int_0^1\sin(\pi x)\log\left(\frac\pi{\sin(\pi x)}\right)\,\mathrm{d}x\tag{4}\\ &=\frac1{2\pi}\int_0^\pi\sin(x)\log(\pi)\,\mathrm{d}x -\frac1{2\pi}\int_0^\pi\sin(x)\log(\sin(x))\,\mathrm{d}x\tag{5}\\ &=\frac{\log(\pi)}\pi-\frac{\log(2)-1}\pi\tag{6}\\[3pt] &=\frac{\log(e\pi/2)}\pi\tag{7} \end{align} $$ Explanation:
$(3)$: substitute $x\mapsto1-x$
$(4)$: average $(2)$ and $(3)$ and use Euler's Reflection Formula
$(5)$: substitute $x\mapsto x/\pi$
$(6)$: apply $(1)$
$(7)$: algebra