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add appendix/common-knowledge
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appendix.typ

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#import "preset.typ":preset-appendix
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#show:preset-appendix
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= 积分常见结论
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== 基本积分表(请背)
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#include "appendix/common-knowledge.typ"

appendix/common-knowledge.typ

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#import "../utils.typ":ee
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+ $integral x^a dif x=1/(a+1)x^(a+1)+C quad (a!=1)$
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#import "../math.typ":*
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== 基本积分表
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以下基本积分是读者应知应会的。其中标红色者为重中之重,必背。
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+ $integral x^a dif x=1/(a+1)x^(a+1)+C space (a!=1)$
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+ $integral 1/x dif x=log abs(x)+C$
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+ $integral ee^x dif x=ee^x+C$
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+ $integral a^x dif x=1/(log a) a^x+C$
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+ $integral sin x dif x=-cos x+C$
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+ $integral cos x dif x=sin x+C$
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+ $integral csc^2x dif x=-cot x+C$
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+ $integral sec^2x dif x=tan x+C$
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+ $integral tan x sec x dif x=sec x+C$
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+ $integral cot x csc x dif x=-csc x+C$
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+ $integral tan x dif x=-log abs(cos x)+C$
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+ $integral cot x dif x=log abs(sin x)+C$
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+ $integral csc x dif x=log abs(tan x/2)+C=-artanh cos x+C$
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+ $integral sec x dif x=log abs(tan(x/2+pi/4))+C=log abs(sec x+tan x)+C=artanh sin x+C$
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+ $integral (dif x)/(1+x^2)=arctan x+C=-arccot x+C_1$
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+ $integral (dif x)/(1-x^2)=artanh x+C=1/2 log abs((1+x)/(1-x))+C$
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+ $integral (dif x)/sqrt(x^2+1)=arsinh x+C=log(x+sqrt(x^2+1))+C$
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+ $integral (dif x)/sqrt(x^2-1)=arcosh abs(x) sgn x+C=log abs(x+sqrt(x^2-1))+C$
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== 常用恒等变换
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以下恒等变换常见于各类积分问题中,足以帮助读者解决绝大多数问题。对于标蓝色者,建议读者记忆;对于其它变换,也建议读者能够熟练推导。
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+ $sqrt(x^2)=abs(x)=x sgn x$
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+ $x=abs(x) sgn x$
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+ $x^3+1=(x+1)(x^2-x+1)$
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+ $x^3-1=(x-1)(x^2+x+1)$
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+ $sin^2x+cos^2x=sec^2x-tan^2x=csc^2x-cot^2x=1$
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+ $sin(a+b)=sin a cos b+cos a sin b$
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+ $sin(a-b)=sin a cos b-cos a sin b$
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+ $cos(a+b)=cos a cos b-sin a sin b$
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+ $cos(a-b)=cos a cos b+sin a sin b$
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+ $sin 2x=2sin x cos x=(2tan x)/(tan^2x+1)$
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+ $cos 2x=cos^2x-sin^2x=1-2sin^2x=2cos^2x-1=(1-tan^2x)/(1+tan^2x)$
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+ $a sin x+b cos x=sqrt(a^2+b^2)sin(x+arctan b/a) space (a>0)$
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+ $1+sin x=(sin x/2+cos x/2)^2$
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+ $1-sin x=(sin x/2-cos x/2)^2$
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+ $1+cos x=2cos^2x/2$
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+ $1-cos x=2sin^2x/2$
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+ $sin^2x=(1-cos 2x)/2$
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+ $cos^2x=(1+cos 2x)/2$
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+ $sin a cos b=(sin(a+b)+sin(a-b))/2$
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+ $cos a sin b=(sin(a+b)-sin(a-b))/2$
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+ $cos a cos b=(cos(a+b)+cos(a-b))/2$
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+ $sin a sin b=(cos(a+b)-cos(a-b))/2$
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+ $sin a+sin b=2sin (a+b)/2 cos (a-b)/2$
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+ $sin a-sin b=2cos (a+b)/2 sin (a-b)/2$
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+ $cos a+cos b=2cos (a+b)/2 cos (a-b)/2$
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+ $cos a-cos b=-2sin (a+b)/2 sin (a-b)/2$
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+ $sin x cos x=((sin x+cos x)^2-1)/2=(1-(sin x+cos x)^2)/2$
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+ $(a sin x+b cos x)^2+(b sin x-a cos x)^2=a^2+b^2$
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== 通题通解
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我们知道,有些形式的积分是必定有解的。在研究一个积分时,我们只需要把原积分化成诸如此的形式,知道它们必定有解足矣。但它们的原函数究竟是多少,我们可不想反复应用那些熟知又麻烦的通法了。#parbreak()
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下面的几个积分是对基本积分表的扩充。你不必记忆,但应当掌握它们的解法。
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+ 记$Y=a x^2+b x+c space(a!=0), Delta=b^2-4a c, x_1, x_2=(-b plus.minus sqrt(Delta))/(2a)$,则$
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integral (dif x)/Y=cases(
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2/sqrt(-Delta)arctan Y'/sqrt(-Delta)+C\,&Delta<0\,,
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-1/(a x+b)+C\,&Delta=0\,,
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1/(a(x_1-x_2))log abs((x-x_1)/(x-x_2))+C\,&Delta>0.
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)
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$
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+ 记$Y=a x^2+b x+c space(a!=0), Delta=b^2-4a c, x_1, x_2=(-b plus.minus sqrt(Delta))/(2a)$,则$
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integral (dif x)/sqrt(Y)=cases(
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1/sqrt(a)log abs(Y'+2sqrt(a Y))+C\,a>0\,,
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1/sqrt(-a)arcsin (-Y')/sqrt(Delta)+C\,a<0.
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)
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$
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+ $integral (dif x)/(a x^n+b)^((n+1)/n)=x/(b root(n,a x^n+b))+C space(n in NN_+)$
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== 常见的非初等积分
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以下积分的原函数是非初等函数。当它们单独存在时,我们不可能得到初等函数的结果。话虽如此,我们仍有可能通过应用分部积分法,抵消两个非初等积分并得到初等函数的结果。
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+ $integral (sin x)/x dif x=op("Si")(x)+C$
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+ $integral (cos x)/x dif x=op("Ci")(x)+C$
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+ $integral (sinh x)/x dif x=op("Shi")(x)+C$
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+ $integral (cosh x)/x dif x=op("Chi")(x)+C$
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+ $integral (ee^x)/x dif x=op("Ei")(x)+C$
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+ $integral (dif x)/(log x)=op("li")(x)+C$
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+ $integral sin x^2 dif x=sqrt(pi/2)S(sqrt(2/pi)x)+C$
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+ $integral cos x^2 dif x=sqrt(pi/2)C(sqrt(2/pi)x)+C_1$#text(fill:gray)[注#footnote[这里的$C$是菲涅耳积分函数,而$C_1$是任意常数。]]
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+ $integral ee^x^2 dif x=sqrt(pi)/2op("erfi")(x)+C$
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+ $integral_0^phi (dif theta)/sqrt(1-k^2sin^2theta)=F(sin phi;k)$
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+ $integral_0_x (dif t)/sqrt((1-t^2)(1-k^2t^2))=F(x;k)$
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+ $integral_0_phi sqrt(1-k^2sin^2theta)dif theta=E(phi;k)$
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+ $integral_0^x sqrt(1-k^2t^2)/sqrt(1-t^2)dif t=E(x;k)$
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+ $integral_0^phi t^(p-1)(1-t)^(q-1)dif t=B(x;p,q)$

collection/level-1.typ

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#import "../utils.typ":question,comment,subst,ref
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#import "../utils.typ":ee
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#import "../math.typ":ee
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#question(
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tag:"(x+1)^3",
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category:red,

collection/level-2.typ

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#import "../utils.typ":question,comment,subst,ref
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#import "../utils.typ":ee
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#import "../math.typ":ee
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#question(
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tag:"cot^2xtan((1+xtanx)/tanx)",
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category:blue,

math.typ

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#let ee=math.upright($e$)
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#let ii=math.upright($i$)
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#let sgn=math.op("sgn")
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#let arccot=math.op("arccot")
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#let arsinh=math.op("arsinh")
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#let arcosh=math.op("arcosh")
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#let artanh=math.op("artanh")

preset.typ

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}
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#let preset-appendix(body)={
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show:preset
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set par(
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first-line-indent:(
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amount:2em,
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all:true,
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)
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)
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show heading.where(level:1):it=>{
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set text(
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size:21pt,

utils.typ

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#let ee=math.upright($e$)
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#let ii=math.upright($i$)
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#let question-id=counter("question")
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#show ref: it=>{
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let target=query(it.target).first()

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