From 2b76ef061aae6b3cd492591e445c9e6452bd357c Mon Sep 17 00:00:00 2001
From: syn <isaqtm@gmail.com>
Date: Tue, 4 Feb 2020 10:43:52 +0300
Subject: Add some hw's

---
 sol0127.tex | 268 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 268 insertions(+)
 create mode 100644 sol0127.tex

(limited to 'sol0127.tex')

diff --git a/sol0127.tex b/sol0127.tex
new file mode 100644
index 0000000..1098687
--- /dev/null
+++ b/sol0127.tex
@@ -0,0 +1,268 @@
+\documentclass[10pt,a5paper]{article}
+\usepackage[svgnames, rgb]{xcolor}
+
+\input{intro}
+
+\lhead{\color{gray} Шарафатдинов Камиль 192}
+\rhead{\color{gray} \texttt{sol0127}}
+\title{ДЗ на 03.02}
+\author{Шарафатдинов Камиль БПМИ-192}
+\date{билд: \today}
+
+
+% -- Here bet  dragons --
+\begin{document}\thispagestyle{empty}
+
+\maketitle
+\clearpage
+\setcounter{page}{1}
+
+%\question{8.a}{
+%    \[
+%        \int \frac{2\sin^3 x + \cos^2 x \sin 2x}{\sin^4 x + 3 \cos^4 x} \dif x = \todo + C
+%    \]
+%}
+
+    
+
+\question{8.c}{
+    \[
+        \int \frac{\dif x}{\cosh^3 x + 3\cosh x} = \frac{
+            2\arctan \sinh x - \arctan \frac{\sinh x}{2}
+        }{6} + C
+    \]
+}
+    \begin{align*}
+        \int \frac{\dif x}{\cosh^3 x + 3\cosh x}
+        &= \int \frac{\frac{\dif \sinh x}{\cosh x}}{\cosh^3 x + 3\cosh x}
+            &\explain{
+                \dif \sinh x = \cosh x \dif x
+            } \\[8pt]
+        &= \int \frac{\dif \sinh x}{\cosh^4 x + 3\cosh^2 x} \\[8pt]
+        &= \int \frac{\dif \sinh x}{\br{ 1 + \sinh^2 x }^2 + 3\br {1 + \sinh^2 x}}
+            &\explain{
+                \cosh^2 x - \sinh^2 x = 1
+            } \\[8pt]
+        &= \int \frac{\dif u}{\br{ 1 + u^2 }^2 + 3\br {1 + u^2}}
+            &\explain{
+                u = \sinh x
+            } \\[8pt]
+        &= \int \frac{\dif u}{4 + 5u^2 + u^4} \\[8pt]
+        &= \int \frac{\dif u}{\br{ u^2 + 1 } \br{ u^2 + 4 }} \\[8pt]
+        &= \int \frac{\dif u}{3} \br{
+            \frac{1}{ u^2 + 1 } - \frac{1}{ u^2 + 4 }
+        } \\[8pt]
+        &= \frac{1}{3} \br{
+            \int \frac{du}{u^2 + 1} - \int \frac{du}{u^2 + 4}
+        } \\[8pt]
+        &= \frac{1}{3} \br{
+            \arctan u - \frac{1}{2}\arctan \frac{u}{2}
+        } + C
+            &\explain{
+                u = \sinh x
+            }\\[8pt]
+        &= \frac{1}{6} \br{
+            2\arctan \sinh x - \arctan \frac{\sinh x}{2}
+        } + C
+    \end{align*}
+
+\clearpage
+\question{8.e}{
+    \[
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x} = \frac{\sqrt{2}}{2} \br{
+            \arctan \br{ \sqrt{2} \tan x + 1 } + \arctan  \br{ \sqrt{2} \tan x - 1 }
+        } + C
+    \]
+}
+
+    \[
+        \sin^4 x + \cos^4 x =
+        \br{ 1 - \cos^2 x }^2 + \cos^4 x =
+        1 - 2\cos^2 x + 2\cos^4 x
+    \]
+    \[
+        \dif x = \cos^2 x \dif\ (\tan x)
+    \]
+    \[
+        1 + \tan^2 x = \frac{1}{\cos^2 x}
+    \]
+
+    \begin{align*}
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x}
+            &= \int \frac{\cos^2 x \dif\ (\tan x)}{1 - 2\cos^2 x + 2\cos^4 x} \\[8pt]
+            &= \int \frac{\dif\ (\tan x)}{\frac{1}{\cos^2 x} - 2 + 2\cos^2 x} \\[8pt]
+            &= \int \frac{\dif\ (\tan x)}{1 + \tan^2 x - 2 + \frac{2}{1 + \tan^2 x}} \\[8pt]
+            &= \int \frac{\dif u}{-1 + u^2 + \frac{2}{1 + u^2}} & [u = \tan x]\\[8pt]
+            &= \int \frac{(1 + u^2) \dif u}{u^4 + 1} \\[8pt]
+    \end{align*}
+
+    По прошлой домашке мы знаем, что
+    \[
+        x^4 + 1 = (x^2 - \sqrt{2} x + 1)(x^2 + \sqrt{2} x + 1)
+    \]
+
+    Разложим на слагаемые:
+    \[
+        \frac{x^2 + 1}{x^4 + 1} =
+        \frac{1}{2} \br{ \frac{1}{x^2 + \sqrt{2} x + 1} + \frac{1}{x^2 - \sqrt{2} + 1} }
+    \]
+
+    \begin{align*}
+        \int \frac{1}{u^2 + \sqrt{2} u + 1}\dif u
+            &= \int \frac{1}{\br{ u + \frac{\sqrt{2}}{2} }^2 + \frac{1}{2}}\dif u \\[8pt]
+            &= \sqrt{2} \arctan \br{ \sqrt{2} u + 1 } + C_1 \\[16pt]
+        \int \frac{1}{u^2 - \sqrt{2} u + 1}\dif u
+            &= \int \frac{1}{\br{ u - \frac{\sqrt{2}}{2} }^2 + \frac{1}{2}}\dif u \\[8pt]
+            &= \sqrt{2} \arctan \br{ \sqrt{2} u - 1 } + C_2 \\[8pt]
+    \end{align*}
+
+    \begin{gather*}
+        \int \frac{\dif x}{\sin^4 x + \cos^4 x} =
+        \int \frac{(\tan^2 x + 1) \dif\ \tan x}{\tan^4 x + 1} =\\[8pt] =
+        \frac{\sqrt{2}}{2} \br{
+            \arctan \br{ \sqrt{2} \tan x + 1 } + \arctan  \br{ \sqrt{2} \tan x - 1 }
+        } + C
+    \end{gather*}
+
+\question{8.g}{
+    \[
+        \int \frac{\dif x}{a \sin x + b \cos x + c}, \qquad c > \sqrt{a^2 + b^2}
+    \]
+}
+
+    Найдем такой интеграл в предположении $a > 1$:
+
+    \begin{align*}
+        \int \frac{\dif x}{\sin x + a}
+        &= \int \frac{\frac{2 \dif u}{1 + u^2}}{\frac{2u}{1 + u^2} + a}
+            &\explain{
+                \displaystyle u = \tan \frac{x}{2}\\[8pt]
+                \displaystyle \dif x = \frac{2 \dif u}{1 + u^2}\\[8pt]
+                \displaystyle \sin x = \frac{2u}{1 + u^2}
+            } \\[8pt]
+        &= \int \frac{2\dif u}{2u + a + au^2} \\[8pt]
+        &= \int \frac{2\dif u}
+            {\br{ \sqrt{a} u + \frac{1}{\sqrt{a}} }^2 + 1 - \frac{1}{a}} \\[8pt]
+        &= \int \frac{2\dif u}
+            {\br{ \sqrt{a} u + \frac{1}{\sqrt{a}} }^2 + \frac{a^2 - 1}{a}} \\[8pt]
+        &= \frac{2}{\sqrt{a}}\int \frac{\dif v}{v^2 + \frac{a^2 - 1}{a}}
+            &\explain{
+                v = \sqrt{a} u + \frac{1}{\sqrt{a}}\\
+                \dif v = \sqrt{a} \dif u
+            } \\[8pt]
+        &= \frac{2\sqrt{a}}{\sqrt{a}\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{\sqrt{a} v}{\sqrt{a^2 - 1}} } + C \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{\sqrt{a} v}{\sqrt{a^2 - 1}} } + C \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{au + 1}{\sqrt{a^2 - 1}} } + C
+            &\explain{
+                v = \sqrt{a} u + \frac{1}{\sqrt{a}}
+            } \\[8pt]
+        &= \frac{2}{\sqrt{a^2 - 1}}
+            \arctan \br{ \frac{a \tan \frac{x}{2} + 1}{\sqrt{a^2 - 1}} } + C
+    \end{align*}
+
+    Теперь, непосредственно задание
+
+    \begin{align*}
+        \int \frac{\dif x}{a \sin x + b \cos x + c}
+        &= \int \frac{\dif x}{r \br{ \frac{a}{r} \sin x + \frac{b}{r} \cos x} + c }
+            &\explain{
+                \displaystyle r = \sqrt{a^2 + b^2}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{\cos \phi \sin x + \sin \phi \cos x + c/r}
+            &\explain{
+                \displaystyle \phi = \arccos \frac{a}{r} = \arcsin \frac{b}{r}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{\sin \br{ \phi + x } + c/r} \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif u}{\sin \br{ u } + c/r}
+            &\explain{
+                u = \phi + x\\
+                du = dx
+            } \\[8pt]
+        &= \frac{2}{r\sqrt{\dfrac{c^2}{r^2} - 1}}
+            \arctan \br{
+                \frac
+                    {\displaystyle \frac{c}{r} \tan \frac{u}{2} + 1}
+                    {\displaystyle \sqrt{\frac{c^2}{r^2} - 1}}
+            } + C
+            &\explain{\text{По доказанному}}\\[8pt]
+        &= \frac{2}{\sqrt{c^2 - r^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle \frac{c}{r} \tan \frac{\arccos \dfrac{a}{r}}{2} + 1}
+                    {\displaystyle \sqrt{\frac{c^2}{r^2} - 1}}
+            } + C \\[8pt]
+        &= \frac{2}{\sqrt{c^2 - r^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle c \tan \frac{\arccos \dfrac{a}{r}}{2} + r}
+                    {\displaystyle \sqrt{c^2 - r^2}}
+            } + C \\[8pt]
+        &= \frac{2}{\sqrt{c^2 - a^2 - b^2}}
+            \arctan \br{
+                \frac
+                    {\displaystyle c \tan \frac{\arccos \dfrac{a}{\sqrt{a^2 + b^2}}}{2} + \sqrt{a^2 + b^2}}
+                    {\displaystyle \sqrt{c^2 - a^2 - b^2}}
+            } + C \\[8pt]
+    \end{align*}
+
+\clearpage
+
+\question{10}{
+    \[
+        \int \frac{\dif x}{(a\sin x + b\cos x)^n}
+    \]
+}
+
+    Найдем рекуррентную формулу для следующего интеграла:
+    \begin{align*}
+        \int \frac{\dif x}{\sin^n x}
+        &= -\int \frac{\dif \cos x}{\sin^{n + 1} x}\\[6pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            \int \br{ \frac{1}{\sin^{n + 1} x} }' \cos x \dif x
+            &\explain{
+                \displaystyle \int Fg \dif x = FG - \int fG \dif x\\
+                \displaystyle F = \frac{1}{\sin^{n + 1} x}\\[10pt]
+                \displaystyle g = \frac{\dif \cos x}{\dif x}\\
+                \displaystyle G = \cos x
+            } \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{\cos^2 x \dif x}{\sin^{n + 2} x} \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{(1 - \sin^2 x) \dif x}{\sin^{n + 2} x} \\[8pt]
+        &= -\frac{\cos x}{\sin^{n + 1} x} -
+            (n + 1) \int \frac{\dif x}{\sin^{n + 2} x} +
+            (n + 1) \int \frac{\dif x}{\sin^n x}
+    \end{align*}
+
+    Пусть $\displaystyle J_n = \int \frac{\dif x}{\sin^n x}$.
+
+    Переобозначим $n = n + 2$ в полученном интеграле, чтобы формула получилась красивой
+    \begin{align*}
+        J_{n - 2} &= -\frac{\cos x}{\sin^{n - 1} x} - (n - 1) J_n + (n - 1) J_{n - 2}\\[8pt]
+        (n - 1)J_n &= -\frac{\cos x}{\sin^{n - 1} x} + (n - 2)J_{n - 2}\\[8pt]
+        J_n &= \frac{\cos x}{(1 - n) \sin^{n - 1} x} + \frac{n - 2}{n - 1}J_{n - 2}
+    \end{align*}
+
+    Тогда:
+    \begin{align*}
+        I_n = \int \frac{\dif x}{(a\sin x + b\cos x)^n}
+        &= \frac{1}{r} \int \frac{\dif x}{(\frac{a}{r}\sin x + \frac{b}{r}\cos x)^n}
+            &\explain{
+                \displaystyle r = \sqrt{a^2 + b^2}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{(\cos \phi \sin x + \sin \phi \cos x)^n}
+            &\explain{
+                \displaystyle \phi = \arccos \frac{a}{r} = \arcsin \frac{b}{r}
+            } \\[8pt]
+        &= \frac{1}{r} \int \frac{\dif x}{(\sin (\phi + x))^n} \\[8pt]
+        &= \frac{\cos (\phi + x)}{r(1 - n) \sin^{n - 1} (\phi + x)} +
+            \frac{n - 2}{n - 1} I_{n - 2} \\[8pt]
+        &= \frac{\cos (\arccos{\frac{a}{\sqrt{a^2 + b^2}}} + x)}
+                {\sqrt{a^2 + b^2}(1 - n) \sin^{n - 1} (\arccos{\frac{a}{\sqrt{a^2 + b^2}}} + x)} +
+            \frac{n - 2}{n - 1} I_{n - 2}
+    \end{align*}
+
+\end{document}
-- 
cgit v1.2.1-18-gbd029

