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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}