path: root/sol0203.tex

\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}


\newcommand{\deft}{\texttt{\\deft is undefined}}
    

\question{1.b}{
    \[
        \int \frac{\dif x}{\sqrt[6]{(x - 7)^7(x - 5)^5}}
        = -3\sqrt[6]{\dfrac{x - 5}{x - 7}} + C
    \]
}

    \begin{align*}
        \int \frac{\dif x}{\sqrt[6]{(x - 7)^7(x - 5)^5}}
            &= -6 \int \frac{t^7 \dif t}{t^7 \br{ \sqrt[6]{x - 5} }^5}
                &\explain{
                    \displaystyle t = \sqrt[6]{\frac{1}{x - 7}}\\[8pt]
                    \displaystyle \frac{\dif t}{\dif x}
                        = -\frac{1}{6} \br{ \sqrt[6]{\frac{1}{x - 7}} }^7
                        = -\frac{1}{6} t^7
                }\\[8pt]
            &= -6 \int \frac{\dif t}{\br{ \sqrt[6]{x - 5} }^5}\\[8pt]
            &= -6 \int \frac{t^5 \dif t}{\br{ \sqrt[6]{1 + 2t^6} }^5}
                &\explain{
                    \dfrac{1}{\sqrt[6]{x - 5}}
                        &= \displaystyle \sqrt[6]{\frac{1}{\frac{1}{t^6} + 2}}\\[8pt]
                        &= \dfrac{t}{\sqrt[6]{1 + 2t^6}}
                }\\[8pt]
            &= 3 \int \frac{u^5 \dif u}{u^7}
             = 3 \int \frac{\dif u}{u^2}
                &\explain{
                    \displaystyle u = \dfrac{1}{\sqrt[6]{1 + 2t^6}}\\
                    \displaystyle \frac{\dif u}{\dif t} = -2t^5 u^7\\
                }\\
            &= -\frac{3}{u} + C\\
            &= -3\sqrt[6]{1 + 2t^6} + C\\
            &= -3\sqrt[6]{1 + \dfrac{2}{x - 7}} + C\\
            &= -3\sqrt[6]{\dfrac{x - 5}{x - 7}} + C
    \end{align*}

\clearpage

\question{7.c}{
    \[
        \int \frac{\dif x}{1 + \sqrt{1 - 2x - x^2}}
    \]
}

    Воспользуемся почти подстановкой Эйлера:
    $
        \displaystyle \sqrt{ax^2 + bx + c} = xt - \sqrt{c}
    $

    \begin{align*}
        \int \frac{\dif x}{1 + \sqrt{1 - 2x - x^2}}
            &= \int \frac{\dif x}{xt}
                &\explain{
                    \displaystyle \sqrt{1 - 2x - x^2} = xt - 1\\[4pt]
                    \displaystyle 1 - 2x - x^2 = x^2t^2 - 2xt + 1\\[4pt]
                    \displaystyle x \br{ t^2 + 1 } = t - 1\\[4pt]
                    \displaystyle x = \dfrac{t - 1}{t^2 + 1}\\[16pt]
                    \displaystyle \dfrac{\dif x}{\dif t} = -2 \dfrac{t^2 - 2t - 1}{(t^2 + 1)^2}
                }\\
            &= \int \frac{-2\dfrac{t^2 - 2t - 1}{(t^2 + 1)^2} \dif t}
                         {2\dfrac{t - 1}{t^2 + 1} t}\\[8pt]
            &= -\int \frac{(t^2 - 2t - 1)(t^2 + 1) \dif t}
                          {(t - 1)(t^2 + 1)^2 t}\\[8pt]
            &= -\int \frac{(t^2 - 2t - 1) \dif t}
                          {(t - 1)(t^2 + 1) t}\\[8pt]
            &= -\int \br{
                \frac{2}{t^2 + 1} + \frac{1}{t} - \frac{1}{t - 1}
            } \dif t\\[8pt]
            &= -2\arctan t - \log{t} + \log(t - 1) + C\\[8pt]
            &= -2\arctan \dfrac{\sqrt{1 - 2x - x^2} + 1}{x} + \log \dfrac{t - 1}{t} + C\\[8pt]
            &= -2\arctan \dfrac{\sqrt{1 - 2x - x^2} + 1}{x}
                 + \log \dfrac{\sqrt{1 - 2x - x^2} + 1 - x}{\sqrt{1 - 2x - x^2} + 1} + C\\[8pt]
    \end{align*}

\clearpage

\question{10.b}{
    \[
        \int \frac{x^8 \dif x}{\sqrt{x^2 + 1}}
    \]
}

    Лемма (вообще говоря, это задача 9):
    \begin{gather*}
        P \in \mathbb{R}_n[x], \qquad
        Q \in \mathbb{R}_{n - 1}[x], \qquad
        R = \sqrt{ax^2 + bx + c} \implies
        \int \frac{P \dif x}{R} = Q R + \lambda \int \frac{\dif x}{R}\\[16pt]
        \br{ QR + \lambda \int \frac{\dif x}{R}}' =
        Q'R + QR' + \frac{\lambda}{R} =
        \frac{Q'R^2}{R} + \frac{Q(2ax + b)}{2R} + \frac{\lambda}{R} =
        \frac{Q'R^2 + \frac{1}{2} Q(2ax + b) + \lambda}{R}
    \end{gather*}

    Надо бы ещё доказать, что такое $Q$ всегда найдется, но нам достаточно того, что в задаче такой $Q$ есть.

    Тогда по лемме нам надо разложить $x^8$ на слагаемые $Q'(x^2 + 1) + Qx + \lambda$
    для некоторого $Q$.

    Пусть $Q = a_7x^7 + \ldots + a_0, \quad Q' = 7a_7x^6 + \ldots + a_1$

    Получится система линейных уравнений, которую я выписывать не буду, а выпишу сразу ответ:
    \[
        Q = \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x, \quad \lambda = \frac{35}{128}
    \]

    Тогда

    \begin{align*}
        \int \frac{x^8 \dif x}{\sqrt{x^2 + 1}} &= \br{
            \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x
        } \sqrt{x^2 + 1} + \frac{25}{128}\int \frac{\dif x}{\sqrt{x^2 + 1}} \\[8pt]&=
        \br{
            \frac{1}{8}x^7 - \frac{7}{48} x^5 + \frac{35}{192} x^3 - \frac{35}{128} x
        } \sqrt{x^2 + 1} + \frac{25}{128} \log \left|x + \sqrt{x^2 + 1}\right| + C
    \end{align*}

\question{17.b}{
    \[
        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
    \]
}

    \newcommand{\brac}[2]{ \br{ \dfrac{#1}{#2} } }
    \newcommand{\sbrac}[2]{ \br{ \frac{#1}{#2} } }
    \renewcommand{\deft}{\br{ \sqrt{x^2 + x + 1} - x }}

    \begin{gather*}
        \sqrt{x^2 + x + 1} = x + t\\
        x^2 + x + 1 = x^2 + 2xt + t^2
    \end{gather*}
    \begin{align*}
        x &= \frac{t^2 - 1}{1 - 2t}\\
        \dif x &= -\frac{2(t^2 - t + 1)}{(1 - 2t)^2} \dif t\\
        x + t &= \frac{t^2 - t + 1}{2t - 1}\\
        x + 3 &= \frac{t^2 - 6t + 2}{1 - 2t}\\
        x^2 + 1 &= \frac{t^4 + 2t^2 - 4t + 2}{(1 - 2t)^2}
    \end{align*}

    \begin{align*}
        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
            &= -2 \int \frac{(x + 3)(t^2 - t + 1)\dif t}{(x^2 + 1)(x + t)(1 - 2t)^2}\\[8pt]
            &= -2 \int \frac{
                \brac{t^2 - 6t + 2}{1 - 2t}(t^2 - t + 1) \dif t
            }{
                \brac{t^4 + 2t^2 - 4t + 2}{(1 - 2t)^2}
                \brac{t^2 - t + 1}{2t - 1}
                (1 - 2t)^2
            }\\[8pt]
            &= -2 \int - \frac{t^2 - 6t + 2}{t^4 + 2t^2 - 4t + 2} \dif t\\[8pt]
            &= 2 \int \frac{t^2 - 6t + 2}{t^4 + 2t^2 - 4t + 2} \dif t\\[8pt]
            &= \frac{1}{\sqrt{2}} \int \br{
                \frac{2 t - 3 \sqrt2 + 4}{-t^2 + \sqrt2 t + \sqrt2 - 2}
                + \frac{2 t + 3 \sqrt2 + 4}{t^2 + \sqrt2 t + \sqrt2 + 2}
            } \dif t\\[8pt]
    \end{align*}

    \renewcommand{\deft}{t}
    \def\firstdenum{\br{ \deft - \frac{\sqrt2}{2} }^2 + \sbrac{\sqrt{6 - 4\sqrt2}}{2}^2}
    \def\firstpoly{ \deft^2 - \sqrt2 \deft - \sqrt2 + 2 }
    \def\firstsqrt{ \sqrt{6 - 4\sqrt2} }

    \def\seconddenum{ \br{ \deft + \frac{\sqrt2}{2} }^2 + \sbrac{\sqrt{6 + 4\sqrt2}}{2}^2 }
    \def\secondpoly{ \deft^2 + \sqrt2 \deft + \sqrt2 + 2 }
    \def\secondsqrt{ \sqrt{6 + 4\sqrt2}  }

    \begin{align*}
        \int \frac{2 t - 3 \sqrt2 + 4}{-t^2 + \sqrt2 t + \sqrt2 - 2}\dif t
            &= -2 \int \frac{t - \frac{3\sqrt2}{2} + 2}{\firstdenum}\dif t\\[8pt]
            &= -2 \int \frac{t - \frac{\sqrt2}{2}}{\firstdenum} \dif t
               -2 \int \frac{-\sqrt2 + 2}{\firstdenum} \dif t\\[8pt]
            &= - \int \frac{\dif \br{ \firstpoly }}{\firstpoly} \dif t
               +2(\sqrt2 - 2) \frac{2}{\firstsqrt} \arctan \frac{2t - \sqrt2}{\firstsqrt}\\[8pt]
            &= -\log{ \left|\firstpoly\right| } -
                4\arctan \frac{2t - \sqrt2}{\firstsqrt} + C_1
    \end{align*}

    \begin{align*}
        \int \frac{2 t + 3 \sqrt2 + 4}{\secondpoly}\dif t
            &= \int \frac{2t + 3\sqrt2 + 4}{\seconddenum} \dif t\\[8pt]
            &= \int \frac{2t + \sqrt2}{\seconddenum} +
               \int \frac{2\sqrt2 + 4}{\seconddenum}\\[8pt]
            &= \log{ \left|\secondpoly\right| } +
               2(\sqrt2 + 2)\frac{2}{\secondsqrt} \arctan \frac{2t + \sqrt2}{\secondsqrt} + C_2\\[8pt]
            &= \log{ \left|\secondpoly\right| } +
                4\arctan \frac{2t + \sqrt2}{\secondsqrt} + C_2
    \end{align*}

    \begin{align*}
        \int \frac{(x + 3) \dif x}{(x^2 + 1)\sqrt{x^2 + x + 1}}
            &= -\log{ \left|\firstpoly\right| } -
                4\arctan \frac{2t - \sqrt2}{\firstsqrt}\\[8pt]
            &  +\log{ \left|\secondpoly\right| } +
                4\arctan \frac{2t + \sqrt2}{\secondsqrt} + C\\[8pt]
    \end{align*}
    \renewcommand{\deft}{\br{\sqrt{x^2 + x + 1} - x}}
    \begin{align*}
            &= -\log{ \left|\firstpoly\right| } -
                4\arctan \frac{2\deft - \sqrt2}{\firstsqrt}\\[8pt]
            &  +\log{ \left|\secondpoly\right| } +
                4\arctan \frac{2\deft + \sqrt2}{\secondsqrt} + C
    \end{align*}

\clearpage
\question{17.c}{
    \[
        \int \frac{\dif x}{(x^3 - x)\sqrt{x^2 + x + 4}}
    \]
}
    \begin{gather*}
        \sqrt{x^2 + x + 4} = x + t\\
        x^2 + x + 4 = x^2 + 2xt + t^2
    \end{gather*}
    \begin{align*}
        x &= \frac{t^2 - 4}{1 - 2t}\\
        \dif x &= -\frac{2(t^2 - t + 4)}{(1 - 2t)^2} \dif t\\
        x + t &= \frac{t^2 - t + 4}{2t - 1}
    \end{align*}

    \medskip

    \renewcommand{\deft}{\sqrt{x^2 + x + 4} - x}

    \begin{align*}
        \int \frac{\dif x}{(x^3 - x)\sqrt{x^2 + x + 4}}
            &= \int \frac{\dif x}{x(x + 1)(x - 1)\sqrt{x^2 + x + 4}}\\[8pt]
            &= -\int \frac{2(t^2 - t + 4) \dif t}{(1 - 2t)^2 x(x + 1)(x - 1)(x + t)}\\[8pt]
            &= -2\int \frac{(t^2 - t + 4) \dif t}{
                (1 - 2t)^2
                \brac{t^2 - 4}{1 - 2t}
                \brac{t^2 - 5 + 2t}{1 - 2t}
                \brac{t^2 - 3 - 2t}{1 - 2t}
                \brac{t^2 - t + 4}{2t - 1}
            }\\[8pt]
            &= 2 \int \frac{(1 - 2t)^2 \dif t}{
                (t^2 - 4)(t^2 + 2t - 5)(t^2 - 2t - 3)
            }\\[8pt]
            &= 2 \int \frac{(1 - 2t)^2 \dif t}{
                (t - 2)(t + 2)(t - \sqrt{6} + 1)(t + \sqrt{6} + 1)(t + 1)(t - 3)
            }\\[8pt]
            &= 2\int \br{
                -\frac{1}{t^2 - 4}
                -\frac{1}{8(t + 1)}
                +\frac{1}{8(t - 3)}
                -\frac{1}{4\sqrt{6}(t + \sqrt{6} + 1)}
                +\frac{1}{4\sqrt{6}(t - \sqrt{6} + 1)}
            } \dif t\\[8pt]
            &= -\arctan \frac{t}{2}
               -\frac{1}{4}\log |t + 1|
               +\frac{1}{4}\log |t - 3|
               -\frac{1}{2\sqrt{6}}\log |t + \sqrt{6} + 1|
               +\frac{1}{2\sqrt{6}}\log |t - \sqrt{6} + 1|\\[8pt]
            &= -\arctan \frac{\deft}{2}
               -\frac{1}{4}\log |\deft + 1|
               +\frac{1}{4}\log |\deft - 3|\\[8pt]
               &-\frac{1}{2\sqrt{6}}\log |\deft + \sqrt{6} + 1|
               +\frac{1}{2\sqrt{6}}\log |\deft - \sqrt{6} + 1|\\[8pt]
    \end{align*}

\end{document}