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authorsyn <isaqtm@gmail.com>2020-04-15 04:35:30 +0300
committersyn <isaqtm@gmail.com>2020-04-15 04:35:30 +0300
commitf642380d55c66e4e5deaaa6c7cef15f6dbfe36c6 (patch)
tree31ed9377de27678b376668131e0cbf8a8639ce16 /sol0210.tex
parent406cd62e6c18587b2859bf77434527f2ac87027d (diff)
downloadtex2-f642380d55c66e4e5deaaa6c7cef15f6dbfe36c6.tar.gz
Reorganize & alg-1
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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}