KaTeX Demo

求

\int{\frac{1+\sin{x}}{\sin{x}(1+\cos{x})}dx}

解由三角函数知道，

\sin{x}

与

\cos{x}

都可以用

\tan{\frac{x}{2}}

的有理式表示，即

\sin{x} = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}} = \frac{2\tan{\frac{x}{2}}}{\sec^2{\frac{x}{2}}} = \frac{2\tan{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}},

\cos{x} = \cos^2{\frac{x}{2}} - \sin^2{\frac{x}{2}} = \frac{1-\tan^2{\frac{x}{2}}}{\sec^2{\frac{x}{2}}} = \frac{1-\tan^2{\frac{x}{2}}}{1+\tan^2{\frac{x}{2}}}.

如果作变换

u = \tan{\frac{x}{2}} \space \lparen-\pi<x<\pi\rparen^①

，那么

\sin{x} = \frac{2u}{1+u^2}, \space \cos{x} = \frac{1-u^2}{1+u^2},

而

x = 2\arctan{u}

，从而

dx = \frac{2}{1+u^2}du.

于是

\begin{align*} &\int{\frac{1+\sin{x}}{\sin{x(1+\cos{x})}}}dx \newline =&\int{\frac{\Big(1+\frac{2u}{1+u^2}\Big)\frac{2du}{1+u^2}}{\frac{2u}{1+u^2}\Big(1+\frac{1-u^2}{1+u^2}\Big)}} =\frac{1}{2}\int{\Big(u+2+\frac{1}{u}\Big)}du \newline =&\frac{1}{2}\Big(\frac{u^2}{2}+2u+\ln|u|\Big)+C =\frac{1}{4}\tan^2{\frac{x}{2}}+\tan{\frac{x}{2}}+\frac{1}{2}\ln\Big|\tan{\frac{x}{2}}\Big|+C. \end{align*}

本例所用的变量代换

u = \tan{\frac{x}{2}}

对三角函数有理式的积分都可以应用.

①　当

x\in((2k-1)\pi,(2k+1)\pi)

时，作变换

u = \tan{\frac{x-2k\pi}{2}} = \tan{\frac{x}{2}}，x = 2k\pi+2\arctan{u}

，以下所得结果相同.

\begin{align*} &\lim\limits_{x\to0}\frac{\sin{2x}}{x} \newline =& 2\lim\limits_{x\to0}\frac{\sin{2x}}{2x} \newline =& 2 \end{align*}

\begin{align*} &\lim\limits_{x\to1}\frac{x^2-4x+3}{x^2-5x+4} \newline =& \lim\limits_{x\to1}\frac{(x-1)(x-3)}{(x-1)(x-4)} \newline =& \lim\limits_{x\to1}\frac{x-3}{x-4} \newline =& \frac{2}{3} \newline \end{align*}

\begin{align*} &\lim\limits_{x\to0}\frac{\sqrt{x+9}-3}{x} \newline =& \lim\limits_{x\to0}\frac{(\sqrt{x+9}-3)(\sqrt{x+9}+3)}{x(\sqrt{x+9}+3)} \newline =& \lim\limits_{x\to0}\frac{1}{\sqrt{x+9}+3} \newline =& \frac{1}{6} \newline \end{align*}

\begin{align*} &\lim\limits_{x\to\infin}(1+\frac{3}{x})^x \newline =& \lim\limits_{x\to\infin}(1+\frac{1}{\frac{x}{3}})^x \newline =& \lim\limits_{x\to\infin}\bigg[(1+\frac{1}{\frac{x}{3}})^\frac{x}{3}\bigg]^3 \newline =& e^3 \newline \end{align*}

y'=6x^2+5

把 y 看作常量，对 x 求导，得

\frac{\partial z}{\partial x}=2x

把 x 看作常量，对 y 求导，得

\frac{\partial z}{\partial y}=2y

\begin{align*} &\int{(x^2+cos{x})dx} \newline =& \int{x^2dx}+\int{cos{x}dx} \newline =& \frac{1}{3}x^3+sin{x}+c \newline \end{align*}

\begin{align*} &\int{\frac{\ln{x}}{x}dx} \newline =& \int{\ln{xd}(\ln{x})} \newline =& \frac{1}{2}\ln^2{x}+c \newline \end{align*}

\begin{align*} &\int_0^1{e^{2x}dx} \newline =& \frac{1}{2}e^{2x}\big\vert_0^1 \newline =& \frac{1}{2}(e^2-e^0) \newline =& \frac{1}{2}(e^2-1) \newline \end{align*}

解方程组

\begin{cases} y = x^2 \newline x = 2 \newline \end{cases}

得交点

(2,4)

以

x

为积分变量，积分区间为

[0,2]

则所求面积为：

\begin{align*} S=& \int_0^2{x^{2}dx}=(\frac{1}{3}x^3)\big\vert_0^2 \newline =& \frac{8}{3} \end{align*}