proj-assn-tex

\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{multline}
	\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{multline}

%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}

%4.392
%1
\begin{equation}
	\int_0^{\pi/4} \ln (\sin x \cos x) \frac{\sin^{2n} x}{\cos^{2n+2} x} \dx = \frac{1}{2n + 1} \left\lbrack (-1)^{n+1} \frac{\pi}{2} - \ln 2 + \frac{1}{2n+1} + 2 \sum_{k=0}^{n-1} \frac{(-1)^{k-1}}{2n - 2k - 1} \right\rbrack
\end{equation}

%2
\begin{equation}
	\int_0^{\pi/4} \ln (\sin x \cos x) \frac{\sin^{2n-1} x}{\cos^{2n+1} x} \dx = \frac{1}{2n} \left\lbrack (-1)^n \ln 2 - \ln 2 + \frac{1}{2n} + (-1)^n \sum_{k=1}^{n-1} \frac{(-1)^{k}}{k} \right\rbrack
\end{equation}

%4.393
%1
\begin{equation}
	\int_0^{\pi/2} \ln \tan x \sin x \dx = \ln 2
\end{equation}

%2
\begin{equation}
	\int_0^{\pi/2} \ln \tan x \cos x \dx = - \ln 2
\end{equation}

%3
\begin{equation}
	\int_0^{\pi/2} \ln \tan x \sin^2 x \dx = - \int_0^{\pi/2} \ln \tan x \cos^2 x \dx = \frac{\pi}{4}
\end{equation}

%4
\begin{equation}
	\int_0^{\pi/4} \frac{\ln \tan x}{\cos 2x} \dx = - \frac{\pi^2}{8}
\end{equation}

%5
\begin{equation}
	\int_0^{\pi/2} \sin x \ln \cot \frac{x}{2} \dx = \ln 2
\end{equation}

%4.394
%1
\begin{equation} \begin{aligned}
	\int_0^{\pi/2} \frac{\ln \tan x \dx}{1 - 2a \cos 2x + a^2} &= \frac{\pi}{2(1 - a^2)} \ln \frac{1 - a}{1 + a}
	  && \left \lbrack a^2 < 1 \right\rbrack \\
	  &= \frac{\pi}{2(a^2 - 1)} \ln \frac{a - 1}{a + 1}
	  && \left \lbrack a^2 > 1 \right\rbrack
\end{aligned} \end{equation}

%2
\begin{equation} \begin{aligned}
	\int_0^{\pi/2} \frac{\ln \tan x \cos 2x \dx}{1 - 2a \cos 2x + a^2} &= \frac{\pi}{4a} \frac{1+a^2}{1 - a^2} \ln \frac{1 - a}{1 + a}
	  && \left \lbrack a^2 < 1 \right\rbrack \\
	  &= \frac{\pi}{4a} \frac{a^2 + 1}{a^2 - 1} \ln \frac{a - 1}{a + 1}
	  && \left \lbrack a^2 > 1 \right\rbrack
\end{aligned} \end{equation}

\end{document}