不定积分的计算-公式

$$ \begin{align*} \sin(a \pm b)=\sin a\cos b \pm \cos a\sin b \\ \cos(a \pm b)=\cos a\cos b \mp \sin a\sin b \\ \tan(a\pm b)=\frac{\tan a \pm \tan b}{1 \mp \tan a\tan b} \\ \implies\cot(a\pm b)=\frac{\cot a\cot b \mp 1}{\cot a \pm \cot b} \\ \end{align*} $$

$$ \begin{align*} & \sin \alpha x \cos\beta x = \frac{1}{2} [\sin (\alpha x + \beta x) + \sin (\alpha x - \beta x) ] \\ & \cos \alpha x\cos\beta x= \frac{1}{2} [\cos (\alpha x + \beta x) + \cos (\alpha x - \beta x) ] \\ & \sin \alpha x\sin\beta x= -\frac{1}{2} [\cos (\alpha x + \beta x) - \cos (\alpha x - \beta x) ] \\ \end{align*} $$

$$ \begin{align*} & \sin 2x =2 \sin x \cos x \\ & \cos 2x = \cos^2x-\sin^2x=1-2\sin ^2x =2\cos ^2x-1 \\ & \tan 2x = \frac{2\tan x}{1- \tan ^2 x} \\ & \cot 2x = \frac{2\cot ^2 x -1 }{2 \cot x} \end{align*} $$

$$ \begin{align*} & \tan \frac{x}{2}=\frac{2\sin \frac{x}{2}\cos \frac{x}{2}}{2\cos ^{2} \frac{x}{2}}=\frac{\sin x}{1+ \cos x }=\frac{\tan x}{secx+1} \end{align*} $$

$$ \begin{align*} \sin ^2x = \frac{1-\cos{2x}}{2} \\ \cos^2x = \frac{1+\cos{2x}}{2} \\ \end{align*} $$

$$ \begin{align*} & arc \cos x +arc \cos (-x) =\pi \\ & arc \cos x+arc \sin x =\frac{\pi}{2} \\ &| \arctan x-\arctan\left( -\frac{1}{x} \right) | =\frac{\pi}{2} \\ & \arctan x + arc \cot x =\frac{\pi}{2} \\ \\ \\ &\arcsin x \pm \arcsin y =\arcsin(x\sqrt{ 1-y^{2} } \pm y\sqrt{ 1-x^{2} }) \\ &arc \cos x \pm arc \cos y =arc \cos(xy \mp \sqrt{ 1-x^{2} } \sqrt{ 1-y^{2} }) \\ &\arctan x \pm \arctan y =\arctan\left( \frac{x \pm y}{1 \mp xy} \right) \\ &arc \cot x \pm arc \cot y=arc \cot\left( \frac{xy \mp 1}{x \pm y} \right) \\ \\ &下面推导过程； \\ & 如令\arcsin x =a,\arcsin y=b,求a+b=？则先求sin(a+b)， \\ &sin(a+b)=sin(\sin a\cos b+\cos b\sin a) \\ &=sin(\sin(\arcsin x)\cos(\arcsin y)+\cos(\arcsin x)\sin(\arcsin y)) \\ &=x\sqrt{ 1-\sin ^{2} (arc \sin y) }+ y \sqrt{ 1-\sin ^{2} (\arcsin x) } \\ &=x\sqrt{ 1-y^{2} }+y\sqrt{ 1-x^{2} } \\ &\implies a+b=\arcsin(x\sqrt{ 1-y^{2} }+y\sqrt{ 1-x^{2} })=\arcsin x+\arcsin y \\ &其他同理 \\ \end{align*} $$

$$ \begin{align*} \\ 1的等量代换（重要！） \\ 1 & =\sin ^{2}x +\cos ^{2}x \\ \\ 1 \pm sin x & = 1 \pm 2\sin \frac{x}{2} \cos \frac{x}{2} = \left( \sin \frac{x}{2} \pm \cos \frac{x}{2} \right) ^2 \\ \\ sec ^{2}x - \tan ^{2}x & =1 \implies secx +\tan x =\frac{1}{secx -\tan x }\\ \csc ^{2}x -\cot ^{2}x & =1 \implies \csc x +\cot x =\frac{1}{\csc x-\cot x } \\ sec x=\frac{1}{\cos x } & = \frac{1}{1-2 \sin ^{2} \frac{x}{2}} =\frac{1}{1-2 \tan ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}} \\ & =\frac{sec^{2}x}{sec^{2}x-2 \tan ^{2} \frac{x}{2}} = \frac{1+\tan ^{2} \frac{x}{2}} {1-\tan ^{2} \frac{x}{2}} \\ \tan x +sec x & = \frac{2\tan \frac{x}{2}}{1-\tan ^{2} \frac{x}{2}} +\frac{1+\tan ^{2} \frac{x}{2} }{1-\tan ^{2} \frac{x}{2}}=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}} \\ \\ | \tan x +sec x | & = \sqrt{ sec^{2}x + 2secx\tan x +\tan ^{2}x } = \sqrt{ 1+2\tan ^{2}x +2sec x\tan x } \\ &=\sqrt{ 1+2\tan ^{2}x +2 \tan x \cdot \frac{1}{\cos x} } = \sqrt{ \frac{1+\sin ^{2}x+2\sin x }{\cos ^{2}x} } \\ &=\sqrt{ \frac{(1+\sin x)^{2} }{1-\sin ^{2}x}} \\ &= \sqrt{ \frac{1+\sin x}{1-\sin x} }\\\\ 同理：| cscx+\cot x | & = \sqrt{ \frac{1+\cos x}{1-\cos x} } \\ \\ \end{align*} $$

$$ \begin{align*} x^3 - a^3=(x-a)(x^{2}+ax+a^{2}) \\ x^3 + a^3=(x+a)(x^{2}-ax+a^{2}) \end{align*} $$

$$ \begin{align*} 欧拉公式：e^{ ix } & =\cos x+i\sin x \\ 变形：令z=e^{ ix } , 则\frac{1}{z} &= \cos x -i\sin x \\ z+\frac{1}{z} & =2\cos x \\ z-\frac{1}{z} & =2i \cdot \sin x \\\\ 再由棣莫佛公式：z^{n}=(e^{ ix })^{n} & =\cos nx +i\sin nx \\ \implies z^{n}+\frac{1}{z^{n}} & =2\cos nx \\ z^{n}-\frac{1}{z^{n}} & =2i \cdot \sin nx \\ \implies \sin nx & =\frac{z^{n}-\frac{1}{z^{n}}}{2i} =\frac{e^{inx}-e^{-inx}}{2i} \\ \cos nx & =\frac{z^{n}+\frac{1}{z^{n}}}{2} =\frac{e^{inx}+e^{-inx}}{2} \\ \end{align*} $$

$$ \begin{align*} \int x^\alpha \, dx = \begin{cases} \frac{x^{\alpha+1}}{\alpha+1} +C, & \alpha \neq -1 \\ \ln x+C, & \alpha =-1 \\ \end{cases} \\ \int \frac{1}{x^{\beta}} \, dx =\frac{1}{1-\beta} \frac{1}{x^{\beta-1}}+C \end{align*} $$

$$ \int \ln x \, dx= x(\ln x -1) +C $$

$$ \begin{align*} \int a^{x} \, dx & =\frac{a^{x}}{\ln a}+C \\ 特别： \int e^{ x }\, dx & = e^{ x }+C \\ \end{align*} $$

$$ \begin{align*} \int \sin x \ dx & =-\cos x+C \\ \int \cos x \, dx & = \sin x+C \\ \int \tan x \, dx & = -\ln\mid cosx| +C \\ \int \cot x \, dx & =\ln \mid \sin x|+C \\ \\ \\ \int sec^2x \, dx & =\tan x+C \\ \int csc ^2 x \, dx & =-\cot x+C \\ \int secx \tan x \, dx & =secx+C \\ \int \csc x \cot x \, dx & =-\csc x+C \\ \\ 正割，余割 \\ \int \csc x \, dx & = \int \frac{- \csc x(\csc x+\cot x)}{\csc x+\cot x} \, dx =-\ln \mid cscx+\cot x|+C \\ & = ln|\csc x -\cot x|+C\\ \\ 也可 &= -\frac{1}{2} ln \frac{1+\cos x}{1-\cos x } +C\\ 也可&=-\ln | \frac{{1+\cos x}}{\sin x} |+C= -\ln| \frac{{2\cos ^{2} \frac{x}{2}}}{2\sin \frac{x}{2}\cos \frac{x}{2}}|+C =\ln |\tan \frac{x}{2}|+C \\ 也可&= \frac{dx}{\sin x}=\frac{dx}{2\sin \frac{x}{2}\cos \frac{x}{2}} = \frac{{sec ^{2} \frac{x}{2}}}{\tan \frac{x}{2}} d\left( \frac{x}{2} \right) =\ln |\tan \frac{x}{2}|+C \\ \\ \int secx \, dx & = \int \frac{ sec x (sec x+\tan x)}{sec x+\tan x} \, dx = \ln |secx+\tan x|+C \\ & =\ln |\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}| +C\\ \\ 也可& = -\ln| secx-\tan x|+C \\ 也可 & =\frac{1}{2} \ln \frac{1 +\sin x}{1-\sin x} +C\\ \\ 双曲正余弦 \\ \int chx \, d & = shx +C \\ \int shx \, dx & =chx+C \\ \end{align*} $$

$$ \begin{align*} \int \frac{1}{\sqrt{ 1-x^{2} }} \, dx & =\arcsin x+C_{1} \\ & = -arc \cos x +C_{2} \\ ps: arc \cos x+arc \sin x & =\frac{\pi}{2} \\ \\ \int \frac{1}{1+x^{2}} \, dx & =arc \tan x +C_{1} \\ & = -arc \cot x+C_{2} \end{align*} $$

$$ \begin{align*} \int \frac{1}{\sqrt{ a^{2}-x^{2} }} \, dx & = arc \sin \frac{x}{a} +C_{1} \\ & = -arc \cos \frac{x}{a} +C_{2} \\ \int \frac{1}{\sqrt{ x^{2} ± a^{2} }} \, dx & = \ln |x+\sqrt{ x^{2} ± a^{2} }|+C \\ \\ \\ \\ \int \sqrt{ a^{2}-x^{2} } \, dx & \overset{v取x}{=} x\sqrt{ a^{2}-x^{2} } + \int x \cdot \frac{x}{\sqrt{ a^{2}-x^{2} }} \, dx \\ & = x\sqrt{ a^{2}-x^{2} }+ \int \frac{-(a^{2}-x^{2})+a^{2}}{\sqrt{ a^{2}-x^{2} }} \, dx \\ & =x\sqrt{ a^{2}-x^{2} }-\int \sqrt{ a^{2}-x^{2} } \, dx +a^{2}\int \frac{1}{\sqrt{ a^{2}-x^{2} }} \, dx \\ \implies \int \sqrt{ a^{2}-x^{2} } \, dx & =\frac{1}{2} \left( x\sqrt{ a^{2}-x^{2} }+ a^{2}arc \sin \frac{x}{a} \right)+C \\ \\ 同理： \int \sqrt{ x^{2} \pm a^{2} } \, dx & \overset{v取x}{=} \frac{1}{2}(x\sqrt{ x^{2} \pm a^{2} } \pm a^{2} \cdot \ln \mid x+\sqrt{ x^{2} \pm a^{2} }|) +C \\ \end{align*} $$

$$ \begin{align*} \int \frac{1}{a^{2}+x^{2}} \, dx & =\frac{1}{a} arc \tan \frac{x}{a} +C \\ \int \frac{1}{x^{2}-a^{2}} \, dx & =\frac{1}{2a} \ln|{\frac{x-a}{x+a}}|+C_{1} \\ &= -\frac{1}{2a} ln | \frac{x+a}{x-a} |+C_{2} \\ \end{align*} $$

$$ \begin{align*} \int \frac{1}{(x+a)(x+b)} \, dx =\frac{1}{b-a} \int \left( \frac{1}{x+a} - \frac{1}{x+b} \right) \, dx \end{align*} $$

$$ \begin{align*} \int e^{ax}\cos bx \, dx & =\frac{e^{ax}}{a^{2}+b^{2}}(a\cos bx+b\sin bx)+C (ab ≠0) \\ \int e^{ax}\sin bx \, dx & =\frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx) \\ & 记忆： \frac{e^{ax}}{a^{2}+b^{2}}(指数的系数 × 三角函数 - 三角函数的导数) \end{align*} $$

$$ \begin{align*} I(m,n) & =\int \cos ^{m}x \sin ^{n}x \, dx \\ & = \frac{\cos ^{m-1}x\sin ^{n+1}x}{m+n}+\frac{m-1}{m+n}I(m-2,n) (m\geq 2,n\geq 0) \\ &=\frac{\sin ^{n-1} \cos ^{m+1}}{m+n}+\frac{n-1}{m+n}I(m,n-2)(n\geq 2,m\geq 0) \end{align*} $$