Josh的学习笔记之高等数学(Part 1)

1. 常见等价无穷小

sin x x \sin x \sim x tan x x \tan x \sim x arcsin x x \arcsin x \sim x
arctan x x \arctan x \sim x ( 1 cos x ) x 2 2 \left( 1 - \cos x \right) \sim \frac{x^2}{2} ln ( 1 + x ) x \ln (1+x) \sim x
( e x 1 ) x \left( e^x -1 \right) \sim x ( a x 1 ) x ln a \left( a^x - 1 \right) \sim x \ln a ( ( 1 + x ) α 1 ) α x \left( \left( 1+x \right) ^\alpha -1 \right) \sim \alpha x

2. 导数基本公式

一般函数 三角函数 反三角函数 双曲函数
(圆函数)
( C ) = 0 \left( C \right)' = 0 ( sin x ) = cos x \left( \sin x \right)' = \cos x ( arcsin x ) = 1 1 x 2 \left( \arcsin x \right)' = \frac{1}{\sqrt{1-x^2}} ( sinh x ) = ( e x e x 2 ) = cosh x \left( \sinh x \right)' = \left(\frac{e^x-e^{-x}}{2}\right)' = \cosh x
( x α ) = α x α 1 ( α Z ) \left( x^\alpha \right)'=\alpha x^{\alpha-1} \left( \alpha \in \mathbf{Z} \right) ( cos x ) = sin x \left( \cos x \right)' = -\sin x ( arccos x ) = 1 1 x 2 \left( \arccos x \right)' = -\frac{1}{\sqrt{1-x^2}} ( cosh x ) = ( e x + e x 2 ) = sinh x \left( \cosh x \right)' = \left( \frac{e^x+e^{-x}}{2} \right)'=\sinh x
( log a x ) = 1 x ln a \left( \log_ax\right)'=\frac{1}{x\ln a} ( tan x ) = 1 cos 2 x = sec 2 x \left( \tan x \right)' = \frac{1}{\cos^2 x}=\sec^2 x ( arctan x ) = 1 1 + x 2 \left( \arctan x \right)' = \frac{1}{1+x^2} ( a r s i n h x ) = ( ln ( x + x 2 + 1 ) ) = 1 x 2 + 1 \left( \mathrm{arsinh} \, x \right)' = \left(\ln \left( x+\sqrt{x^2 + 1} \right) \right)' = \frac{1}{x^2+1}
( ln ( \ln | x x | ) = 1 x )'=\frac{1}{x} ( cot x ) = 1 sin 2 x = csc 2 x \left( \cot x \right)' = -\frac{1}{\sin^2 x} = -\csc^2 x ( a r c c o t x ) = 1 1 + x 2 \left( \mathrm{arccot} \, x \right)' = -\frac{1}{1+x^2} ( a r c o s h x ) = ( ln ( x + x 2 1 ) ) = 1 x 2 1 \left( \mathrm{arcosh} \, x \right)' = \left( \ln \left( x + \sqrt{x^2 - 1} \right) \right)' = \frac{1}{\sqrt{x^2-1}}
( sec x ) = sec x tan x \left( \sec x \right)' = \sec x \cdot \tan x
( csc x ) = csc x cot x \left( \csc x \right)' = \csc x \cdot \cot x
  • Newton-Leibniz 公式:
    ( u v ) ( n ) = u ( n ) v + n u ( n 1 ) v + n ( n 1 ) 2 ! u ( n 2 ) v + + n ( n 1 ) ( n k + 1 ) k ! u ( n k ) v ( k ) + + u v ( n ) = k = 0 n C n k u ( u k ) v k \begin{aligned} \left( u \cdot v \right)^{\left( n \right)} &= u^{\left( n \right)}v + nu^{\left( n-1 \right)}v' + \frac{n \left( n-1 \right)}{2!}u^{\left( n-2 \right)}v'' + \cdots + \frac{n\left( n-1 \right)\cdots\left(n-k+1\right)}{k!}u^{\left( n-k \right)}v^{(k)} + \cdots + uv^{\left( n \right)} \\ &= \sum_{k=0}^{n}C_n^k u^{\left( u-k \right)}v^k \end{aligned} 可以注意到 Newton-Leibniz 公式具有类似二项式定理展开的形式。

3. 几个初等函数的 Maclaurin 公式

  1. Taylor 公式的一般形式
    f ( x ) = f ( x 0 ) + f ( x 0 ) ( x x 0 ) + f ( x 0 ) 2 ! ( x x 0 ) 2 + + f ( n ) ( x 0 ) n ! ( x x 0 ) n + R n f(x)=f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + R_n 其中
    R n = f ( n + 1 ) ( ξ ) ( n + 1 ) ! ( x x 0 ) n + 1 ( ξ x x 0 ) = o ( ( x x 0 ) n ) ( x x 0 ) R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} \, (\xi\,在\,x\,和\,x_0\,之间) = o\left( \left( x - x_0 \right)^n \right)(x\to x_0)

  2. Maclaurin 公式的一般形式
    f ( x ) = f ( 0 ) + f ( 0 ) x + f ( 0 ) 2 ! x 2 + + f ( n ) ( 0 ) n ! x n + R n f(x)=f(0)+f'(0)\,x + \frac{f''(0)}{2!}x^2 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + R_n 其中
    R n = f ( n + 1 ) ( ξ ) ( n + 1 ) ! x n + 1 ( ξ 0 x ) = f ( n + 1 ) ( θ x ) ( n + 1 ) ! x n + 1 ( 0 < θ < 1 ) = o ( ( x x 0 ) n ) ( x x 0 ) R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} \, (\xi\,在\,0\,和\,x\,之间) = \frac{f^{(n+1)}(\theta x)}{(n+1)!} x^{n+1}(0<\theta<1) = o\left( \left( x - x_0 \right)^n \right)(x\to x_0)

  3. 指数函数的 Maclaurin 展开式
    e x = 1 + x + x 2 2 + x 3 3 + + x n n ! + e θ x ( n + 1 ) ! ( 0 < θ < 1 ) e^x=1+x+\frac{x^2}{2} + \frac{x^3}{3} + \cdots +\frac{x^n}{n!} + \frac{e^{\theta x}}{(n+1)!}(0<\theta<1)

  4. 正弦函数的 Maclaurin 展开式
    sin x = x x 3 3 ! + x 5 5 ! + + ( 1 ) m 1 x 2 m 1 ( 2 m 1 ) ! + R 2 m \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots + (-1)^{m-1}\frac{x^{2m-1}}{(2m-1)!} + R_{2m} 其中
    R 2 m = sin ( θ x + ( 2 m 1 ) π 2 ) ( 2 m + 1 ) ! x 2 m + 1 = o ( x 2 m ) ( 0 < θ < 1 ) R_{2m}=\frac{\sin\left( \theta x + \frac{(2m-1)\pi}{2}\right)}{(2m+1)!}x^{2m+1} = o(x^{2m})(0<\theta <1)

  5. 余弦函数的 Maclaurin 展开式
    cos x = 1 x 2 2 ! + x 4 4 ! + + ( 1 ) m x 2 m 2 m ! + R 2 m + 1 \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots + (-1)^m\frac{x^{2m}}{2m!} + R_{2m+1} 其中 R 2 m + 1 = cos ( θ x + ( m + 1 ) π ) ( 2 m + 2 ) ! x 2 m + 2 = o ( x 2 m + 2 ) ( 0 < θ < 1 ) R_{2m+1}=\frac{\cos(\theta x + (m+1)\pi)}{(2m+2)!} x^{2m+2}=o(x^{2m+2}) (0<\theta <1)

  6. ln ( 1 + x ) \ln (1+x) Maclaurin 展开式
    ln ( 1 + x ) = x x 2 2 + x 3 3 x 4 4 + + ( 1 ) n 1 x n n + R n ( x ) \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} + \cdots + \frac{(-1)^{n-1}x^n}{n} + R_n(x) 其中
    R n ( x ) = ( 1 ) n ( n + 1 ) ( 1 + θ x ) n + 1 x n + 1 = o ( x n ) ( 0 < θ < 1 ) R_n(x) = \frac{(-1)^n}{(n+1)(1+\theta x)^{n+1}}x^{n+1} = o(x^n)(0<\theta <1)

  7. ( 1 + x ) α (1+x)^\alpha Maclaurin 展开式
    ( 1 + x ) α = 1 + α x + α ( α 1 ) 2 ! x 2 + + α ( α 1 ) ( α n + 1 ) n ! x n + R n ( x ) (1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \cdots + \frac{\alpha(\alpha - 1)\cdots (\alpha-n+1)}{n!}x^n + R_n(x) 其中
    R n ( x ) = α ( α 1 ) ( α n ) ( n + 1 ) ! ( 1 + θ x ) α n 1 x n + 1 = o ( x ) ( 0 < θ < 1 ) R_n(x) = \frac{\alpha(\alpha -1)\cdots(\alpha-n)}{(n+1)!}(1+\theta x)^{\alpha-n-1}x^{n+1} = o(x)(0<\theta <1)

  8. 1 1 x \frac{1}{1-x} Maclaurin 展开式
    1 1 x = 1 + x + x 2 + + x n + o ( x n ) \frac{1}{1-x} = 1+x+x^2+\cdots+x^n +o(x^n)


4. 曲率及曲率半径

曲率 曲率半径
K = d α d s = tan α = y α = arctan y y 1 + y 2 d x 1 + y 2 d x = y ( 1 + y ) 3 2 K=\left\| \dfrac{\mathrm{d}\alpha}{\mathrm{d}s}\right\| \xlongequal{\tan \alpha = y' \Rightarrow \alpha = \arctan y'} \dfrac{\left\| \frac{y''}{1+y'^2}\mathrm{d}x \right\|}{\sqrt{1+y'^2}}\mathrm{d}x = \dfrac{\left\| y'' \right\|}{\left( 1 + y' \right)^\frac{3}{2}} R = 1 K R = \frac{1}{K}

注:表中的 \|\cdot\| 代表绝对值含义,此处因为和Markdown的表格分隔符“ | ”冲突而写成该形式。


5. 积分常用公式

描述 公式
1 定积分的定义 0 1 f ( x ) d x = I = lim λ 0 i = 1 n f ( ξ i ) Δ x i λ = max { Δ x 1 , Δ x 2 , , Δ x n } \int^1_0f(x)\mathrm{d}x=I=\lim_{\lambda \to 0}\sum_{i=1}^{n}f(\xi_i)\Delta x_i,其中\lambda = \max \left\{ \Delta x_1,\Delta x_2, \cdots, \Delta x_n \right\}
2 变上限积分 a x f ( x ) d x = a x f ( t ) d t Φ ( x ) = a x f ( t ) d t Φ ( x ) = d d x a x f ( t ) d t = f ( x ) 考察定积分\int^x_af(x)\mathrm{d}x=\int^x_af(t)\mathrm{d}t,对\Phi(x)=\int_a^xf(t)\mathrm{d}t,有\Phi'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\int_a^xf(t)\mathrm{d}t = f(x)
3. d d x α ( x ) β ( x ) f ( t ) d t = f [ β ( x ) ] β ( x ) f [ α ( x ) ] α ( x ) \frac{\mathrm{d}}{\mathrm{d}x}\int^{\beta(x)}_{\alpha(x)}f'(t)\mathrm{d}t = f\left[ \beta\left(x\right)\right]\beta'(x)-f[\alpha(x)]\alpha'(x)
4 上下限对称的积分 f ( x ) a a f ( x ) d x = 2 0 a f ( x ) d x f ( x ) a a f ( x ) d x = 0 若f(x)为偶函数,则\int_{-a}^af(x)\mathrm{d}x=2\int_0^af(x)\mathrm{d}x;若f(x)为奇函数,则\int_{-a}^af(x)\mathrm{d}x=0
5 周期函数的积分 f ( x + T ) = f ( x ) 0 n T f ( x ) d x = n 0 T f ( x ) d x f(x+T)=f(x)\Rightarrow\int_0^{nT}f(x)\mathrm{d}x=n\int^T_0f(x)\mathrm{d}x
6 分部积分法 u v d x = u v u v d x u d v = u v v d u \int uv'\mathrm{d}x=uv - \int u'v\mathrm{d}x \Leftrightarrow \int u\mathrm{d}v = uv - \int v\mathrm{d}u

6. 基本积分表

积分式
1 k d x = k x + C d x = x + C \int k\mathrm{d}x=kx+C,特别地\int\mathrm{d}x=x+C
2 x μ d x = x μ + 1 μ + 1 + C ( μ 1 ) d x x 2 = 1 x + C , d x x = 2 x + C \int x^\mu\mathrm{d}x=\frac{x^{\mu+1}}{\mu+1}+C(\mu\ne-1),特别地\int \frac{\mathrm{d}x}{x^2}=-\frac{1}{x}+C,\int\frac{\mathrm{d}x}{\sqrt{x}}=2\sqrt{x}+C
3 d x x = ln x + C \int\frac{\mathrm{d}x}{x}=\ln \|x\|+C
4 d x 1 + x 2 = arctan x + C \int\frac{\mathrm{d}x}{1+x^2} = \arctan x +C
5 d x 1 x 2 = arcsin x + C \int\frac{\mathrm{d}x}{\sqrt{1-x^2}}=\arcsin x+C
6 sin x d x = cos x + C cos x d x = sin x + C \int\sin x\mathrm{d}x = -\cos x+C,\int\cos x\mathrm{d}x = \sin x +C
7 d x cos 2 x = sec 2 x d x = tan x + C d x sin 2 x = csc 2 x d x = cot x + C \int\frac{\mathrm{d}x}{\cos^2x} = \int\sec^2x\mathrm{d}x = \tan x +C,\int\frac{\mathrm{d}x}{\sin^2x}=\int\csc^2x\mathrm{d}x = -\cot x +C
8 a x d x = a x ln a + C e x d x = e x + C \int a^x\mathrm{d}x = \frac{a^x}{\ln a} + C,特别地\int e^x\mathrm{d}x = e^x +C
9 sinh x d x = cosh x + C cosh x d x = sinh x + C \int \sinh x\mathrm{d}x = \cosh x + C,\int \cosh x \mathrm{d}x=\sinh x +C

注:表中的 \|\cdot\| 代表绝对值含义,此处因为和Markdown的表格分隔符“ | ”冲突而写成该形式。


7. 三角函数相关公式

公式
1 sin 2 x + cos 2 x = 1 , tan 2 x + 1 = sec 2 x , cot 2 x + 1 = csc 2 x \sin^2x+\cos^2x=1, \tan^2x+1=\sec^2x,\cot^2x+1=\csc^2x
2 cosh 2 x sinh 2 x = 1 , sinh ( x ± y ) = sinh x cosh y ± sinh y cosh x , cosh ( x ± y ) = cosh x cosh y , sinh 2 x = 2 sinh x cosh x , cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 x 1 \begin{aligned}&\cosh^2x-\sinh^2x=1,\\ &\sinh(x\pm y)=\sinh x\cdot\cosh y \pm \sinh y\cdot\cosh x, \cosh(x\pm y) = \cosh x\cdot\cosh y,\\ &\sinh2x = 2\sinh x\cdot \cosh x, \cosh 2x = \cosh^2x+\sinh^2x = 2\cosh^2x - 1\end{aligned}
3
积化和差公式
sin α cos β = 1 2 [ sin ( α + β ) + sin ( α β ) ] , cos α sin β = 1 2 [ sin ( α + β ) sin ( α β ) ] , cos α cos β = 1 2 [ c o s ( α + β ) + c o s ( α β ) ] , sin α sin β = 1 2 [ cos ( α + β ) cos ( α β ) ] \sin\alpha\cos\beta=\frac12[\sin(\alpha+\beta)+\sin(\alpha-\beta)],\cos\alpha\sin\beta = \frac12[\sin(\alpha+\beta) - \sin(\alpha-\beta)],\\ \cos\alpha\cos\beta = \frac12[cos(\alpha+\beta) + cos(\alpha - \beta)], \sin\alpha\sin\beta=-\frac12[\cos(\alpha+\beta) - \cos(\alpha-\beta)]
4
和差化积公式
sin α + sin β = 2 sin ( α + β 2 ) cos ( α β 2 ) , sin α sin β = 2 sin ( α β 2 ) cos ( α + β 2 ) , cos α + cos β = 2 cos ( α + β 2 ) cos ( α β 2 ) , cos α cos β = 2 sin ( α + β 2 ) sin ( α β 2 ) \sin\alpha + \sin\beta = 2\sin\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \sin\alpha - \sin\beta = 2\sin\left(\frac{\alpha-\beta}{2}\right) \cos\left(\frac{\alpha+\beta}{2}\right), \\ \cos\alpha + \cos\beta = 2\cos\left(\frac{\alpha+\beta}{2}\right) \cos\left(\frac{\alpha-\beta}{2}\right), \cos\alpha - \cos\beta = -2\sin\left(\frac{\alpha+\beta}{2}\right) \sin\left(\frac{\alpha-\beta}{2}\right)
5
三角代换
a 2 x 2 x = a sin t a cos t , a 2 + x 2 x = a tan t a sec t , x 2 a 2 x = a sec t a tan t ( a > 0 , t ( π 2 , π 2 ) ) \sqrt{a^2-x^2}\xrightarrow{x=a\sin t}a\cos t, \sqrt{a^2+x^2} \xrightarrow{x=a\tan t}a\sec t, \sqrt{x^2-a^2}\xrightarrow{x=a\sec t}a\tan t \left( a>0,t\in\left( -\frac{\pi}{2} , \frac{\pi}{2} \right) \right)
6 a 2 x 2 d x = a 2 2 arcsin x a + 1 2 x a 2 x 2 + C ( a > 0 ) \int\sqrt{a^2-x^2}\mathrm{d}x = \frac{a^2}{2}\arcsin\frac xa + \frac12x\sqrt{a^2-x^2} + C(a>0)
7 x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 ln x + x 2 + a 2 + C ( a > 0 ) \int\sqrt{x^2+a^2}\mathrm{d}x = \frac x2\sqrt{x^2+a^2} + \frac{a^2}2\ln \left\| x+\sqrt{x^2+a^2} \right\|+C(a>0)
8 1 x 2 + a 2 d x = ln x + x 2 + a 2 + C , 1 x 2 a 2 d x = ln x + x 2 + a 2 ( a > 0 ) 1 x 2 + 1 d x = ( x + x 2 + 1 ) = a r s i n h x , 1 x 2 1 d x = ln ( x + x 2 1 ) = a r c o s h x \int\frac{1}{\sqrt{x^2+a^2}}\mathrm{d}x = \ln\left\| x+ \sqrt{x^2+a^2} \right\| +C, \int\frac{1}{\sqrt{x^2-a^2}}\mathrm{d}x = \ln\left\| x + \sqrt{x^2+a^2}\right\|(a>0)\\ 特别地,\int\frac{1}{\sqrt{x^2+1}}\mathrm{d}x=\left(x+\sqrt{x^2+1}\right) = \mathrm{arsinh}\, x,\int\frac{1}{\sqrt{x^2-1}}\mathrm{d}x = \ln\left( x+\sqrt{x^2-1}\right) = \mathrm{arcosh}\, x
9 sec x d x = 1 cos x d x = sec 2 x + sec x tan x sec x + tan x d x = 1 sec x + tan x d ( sec x + tan x ) = ln sec x + tan x + C \int\sec x\mathrm{d}x = \int\frac{1}{\cos x}\mathrm{d}x = \int\frac{\sec^2x+\sec x \tan x}{\sec x + \tan x}\mathrm{d}x = \int\frac{1}{\sec x + \tan x}\mathrm{d}(\sec x + \tan x) = \ln \left\| \sec x + \tan x \right\| + C
10 csc x d x = 1 sin x d x = ln tan x 2 + C = 1 2 ln 1 cos x 1 + cos x + C \int\csc x\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}x = \ln\left\| \tan\frac x2 \right\| + C = \frac 12 \ln \left\| \frac{1-\cos x}{1 + \cos x}\right\| + C
11 tan x d x = sin x cos x d x = 1 cos x d ( cos x ) = ln cos x + C \int\tan x\mathrm{d}x = \int\frac{\sin x}{\cos x}\mathrm{d}x = -\int\frac{1}{\cos x}\mathrm{d}(\cos x) = -\ln \left\| \cos x\right\| +C
12 1 tan x d x = cos x sin x d x = 1 sin x d ( sin x ) = ln sin x + C \int\frac{1}{\tan x}\mathrm{d}x = \int\frac{\cos x}{\sin x}\mathrm{d}x = \int\frac{1}{\sin x}\mathrm{d}(\sin x) = \ln\left\| \sin x \right\| + C
13 arctan x d x = x arctan x ln 1 + x 2 + C \int\arctan x\mathrm{d}x = x\arctan x - \ln \left\| 1+x^2 \right\| +C
14
Wallis 公式
0 π 2 sin n x d x = 0 π 2 cos n x d x = { n 1 n n 3 n 2 1 2 π 2 , n n 1 n n 3 n 2 2 3 1 , n = { ( n 1 ) ! ! n ! ! π 2 , n ( n 1 ) ! ! n ! ! , n \int^{\frac{\pi}{2}}_{0}\sin^nx\mathrm{d}x = \int^{\frac{\pi}{2}}_{0}\cos^nx\mathrm{d}x = \begin{cases} \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac12\cdot\frac\pi2,&n为偶数 \\ \frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdot\cdots\cdot\frac23\cdot1,&n为奇数 \end{cases} = \begin{cases} \frac{(n-1)!!}{n!!}\cdot\frac\pi2,&n为偶数 \\ \frac{(n-1)!!}{n!!},&n为奇数 \end{cases}
15 0 π sin n x d x = 2 0 π 2 sin n x d x , 0 π cos n x d x = { 2 0 π 2 cos n x d x , n 0 , n 0 2 π sin n x = 0 2 π cos n x d x = { 4 0 π 2 sin n x d x , n 0 , n 0 π 2 f ( sin x ) d x = 0 π 2 f ( cos x ) d x , 0 π f ( sin x ) d x 0 π f ( cos x ) d x 0 π π f ( sin x ) d x = π 2 0 π f ( sin x ) d x = π 0 π 2 f ( sin x ) d x \int_0^\pi\sin^nx\mathrm{d}x = 2\int^\frac\pi2_0\sin^nx\mathrm{d}x,\int_0^\pi\cos^nx\mathrm{d}x = \begin{cases} 2\int^\frac\pi2_0\cos^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases} \\ \int_0^{2\pi}\sin^nx = \int_0^{2\pi}\cos^nx\mathrm{d}x = \begin{cases} 4\int^\frac\pi2_0\sin^nx\mathrm{d}x,&n为偶数 \\ 0,&n为奇数 \end{cases} \\ \int_0^\frac\pi2f(\sin x)\mathrm{d}x = \int_0^\frac\pi2f(\cos x)\mathrm{d}x, \int_0^\pi f(\sin x)\mathrm{d}x \ne \int_0^\pi f(\cos x)\mathrm{d}x \\ \int_0^\pi\pi f(\sin x)\mathrm{d}x = \frac\pi2\int_0^\pi f(\sin x)\mathrm{d}x = \pi\int_0^\frac\pi2 f(\sin x)\mathrm{d}x
16
万能公式
u = tan x 2 , x = 2 arctan u sin x = 2 u 1 + u 2 , cos x = 1 u 2 1 + u 2 , d x = 2 1 + u 2 d u 令u=\tan\frac x2,x=2\arctan u,则\sin x = \frac{2u}{1+u^2},\cos x=\frac{1-u^2}{1+u^2}, \mathrm{d}x = \frac{2}{1+u^2}\mathrm{d}u
17 a cos x + b sin x c cos x + d sin x = A d x + B d ( c cos x + d sin x ) c cos x + d sin x ( a cos x + b sin x = A ( c cos x + d sin x ) + B ( c cos x + d sin x ) , A B ) \int \frac{a\cos x+ b\sin x}{c\cos x + d \sin x} = A\int\mathrm{d}x+B\int\frac{\mathrm{d}(c\cos x+d \sin x)}{c \cos x + d \sin x} \\ (设a\cos x + b\sin x = A(c \cos x + d \sin x) + B(c \cos x + d \sin x)',由待定系数法求出A、B)

注:表中的 \|\cdot\| 代表绝对值含义,此处因为和Markdown的表格分隔符“ | ”冲突而写成该形式。


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