\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}