\input amstex
\input cyracc.def
\font\tencyr=wncyr10
\def\cyr#1{{\tencyr\cyracc#1}}

$$\split
\int_0^{2\pi}\frac{\hbox{d} x}{\sin^4 x+\cos^4 x}
&=\int_{0^+}^{(\pi/2)^-}+\int_{(\pi/2)^+}^{\pi^-}
+\int_{\pi^+}^{\bigl((3/2)\pi\bigr)^-}+\int_{\bigl((3/2)
\pi\bigr)^+}^{(2\pi)^-}=\\
\intertext{$[$\cyr{smena}:
 $\tan x=t$, $\hbox{d}x=\frac{\hbox{d}t}{1+t^2}$$]$}
&=\int_{0^+}^{+\infty}+\int_{-\infty}^{0^-}
 +\int_{0^+}^{+\infty}+\int_{-\infty}^{0^-}=\\
&=2\int_{-\infty}^{0^-}+2\int_{0^+}^{+\infty}=\\
\intertext{$[$\cyr{smena}: $t-1/t=m$,$1+1/t^2\hbox{d}t=\hbox{d}m$$]$}
&=2\int_{-\infty}^{+\infty}+2\int_{-\infty}^{+\infty}=\\
&=4\int_{-\infty}^{+\infty}\frac{\hbox{d}m}{2+m^2}=\\
&=2\int_{-\infty}^{0^-}\frac{\sqrt{2}\,\hbox{d}\bigl(\frac{1}{\sqrt{2}}m\bigr)}{1+\bigl(\frac{1}
{\sqrt{2}}m\bigr)^2}+2\int_{0^+}^{+\infty}\frac{\sqrt{2}
\,\hbox{d}\bigl(\frac{1}{\sqrt{2}}m\bigr)}{1+\bigl(\frac{1}
{\sqrt{2}}m\bigr)^2}=\\
&=2\sqrt{2}\arctan\bigl((1/\sqrt{2})m\bigr)\bigr
|_{-\infty}^{0^-}+2\sqrt{2}\arctan\bigl((1/\sqrt{2})m\bigr)
\bigr|_{0^+}^{+\infty}=\\
&=\sqrt{2}\pi+\sqrt{2}\pi=2\sqrt{2}\pi
\endsplit$$
\bye