From f642380d55c66e4e5deaaa6c7cef15f6dbfe36c6 Mon Sep 17 00:00:00 2001
From: syn <isaqtm@gmail.com>
Date: Wed, 15 Apr 2020 04:35:30 +0300
Subject: Reorganize & alg-1

---
 sol0210.tex | 56 --------------------------------------------------------
 1 file changed, 56 deletions(-)
 delete mode 100644 sol0210.tex

(limited to 'sol0210.tex')

diff --git a/sol0210.tex b/sol0210.tex
deleted file mode 100644
index bb4ad94..0000000
--- a/sol0210.tex
+++ /dev/null
@@ -1,56 +0,0 @@
-\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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