rendered paste body\documentclass[a4paper,fleqn]{amsart}
\usepackage{fullpage} % This package makes the margins smaller.
% Here you can define your own macros
\newcommand{\dt}{\,\mathrm{d}t} % good in integrals over t
\newcommand{\dx}{\,\mathrm{d}x} % same for integrals over x
\begin{document}
%4.391
%1
\begin{equation}
\int_0^{\pi/4} (\ln \cos 2x)^n \cos^{p-1} 2x \tan x \dx \\
= \int_0^{\pi/4} (\ln \sin 2x)^n \sin^{p-1} 2x \tan \left( \frac{\pi}{4} - x \right) \! \dx \\
= \frac{1}{2} \beta^{(n)} (p)
\lbrack p > 0 \rbrack
\end{equation}
%2
\begin{equation}
\int_0^{\pi/4} (\ln \sin 2x)^n \sin^{p-1} 2x \tan \left( \frac{\pi}{4} + x \right) \dx = \frac{(-1)^n n!}{2} \, \zeta (n+1,p)
\end{equation}
%3
\begin{equation}
\int_0^{\pi/4} (\ln \cos 2x)^{2n-1} \tan x \dx = \frac{1-2^{2n-1}}{4n} \pi^{2n} |B_{2n}| \\
\lbrack n = 1, 2, \ldots \rbrack
\end{equation}
%4
\begin{equation}
\int_0^{\pi/4} (\ln \cos 2x)^{2n} \tan x \dx = \frac{2^{2n} 1}{2^{2n+1}} (2n)! \, \zeta (2n + 1)
\end{equation}
\end{document}