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\documentclass[10pt,a5paper]{article}
\usepackage[svgnames, rgb]{xcolor}
\input{intro}
\lhead{\color{gray} Шарафатдинов Камиль 192}
\rhead{\color{gray} \texttt{sol0203}}
\title{ДЗ на 10.02}
\author{Шарафатдинов Камиль БПМИ-192}
\date{билд: \today}
% -- Here bet dragons --
\begin{document}\thispagestyle{empty}
\maketitle
\clearpage
\setcounter{page}{1}
\question{1.f}{
\[
\int_1^e \sin \log x \dif x
\]
}
\begingroup
\setlength{\jot}{8pt}
\begin{align*}
I = \int \sin \log \dif x
&= \int e^u \sin u \dif u
&\explain{
\displaystyle u = \log x\\
\displaystyle \dif u = \frac{\dif x}{x} = \frac{\dif x}{e^u}
}\\
&= - e^u \cos u + \int e^u \cos u \dif u\\
&= - e^u \cos u + e^u \sin u - \int e^u \sin u \dif u
\end{align*}
\endgroup
\begin{flalign*}
2I = e^u (\sin u - \cos u) + C\\
I = \frac{e^u}{2} (\sin u - \cos u) + C =
\frac{x}{2} (\sin \log x - \cos \log x) + C
\end{flalign*}
\[
\int_1^e \sin \log x \dif x =
\frac{e}{2} (\sin 1 - \cos 1) - \frac{1}{2} (0 - 1) =
\frac{e}{2} (\sin 1 - \cos 1) + \frac{1}{2}
\]
\vspace*{\fill}
P.S. Ну проверь хотя бы одну задачу, пожаааалуйста
\end{document}
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