一些高数积分笔记

Last updated on February 29, 2024 8:33 PM

救救期中。。

三角函数的一些公式

tan2x+1=sec2x\tan^{2}x+1=\sec^{2}x 是很重要的关于 tanx\tan x 的处理,很多时候和 dtanx=sec2xdx\mathrm{d}\tan x = \sec^{2} x \mathrm{d}x 一起处理问题。

然后是三角函数的 Reduction Formula,采用分部积分来处理。

sinmxdx=sinm1xdcosx=cosxdsinm1xcosxsinm1x=(m1)sinm2xcos2xdxcosxsinm1x=(m1)sinm2x(1sin2x)dxcosxsinm1xmsinmxdx=(m1)sinm2xdxsinm1xcosxsinmxdx=m1msinm2xdx1msinm1xcosx\begin{aligned} \int\sin^m x \mathrm{d}x =&-\int \sin^{m-1}x\mathrm{d}\cos x\\ =&\int \cos x \mathrm{d} \sin^{m-1}x - \cos x \sin^{m-1}x\\ =&(m-1)\int\sin^{m-2}x \cos^{2}x\mathrm{d}x - \cos x \sin^{m-1}x\\ =&(m-1)\int\sin^{m-2}x(1-\sin^{2}x)\mathrm{d}x - \cos x\sin^{m-1}x\\ m \int \sin^m x \mathrm{d}x=& (m-1)\int \sin^{m-2}x\mathrm{d}x-\sin^{m-1}x\cos x\\ \int \sin^m x\mathrm{d}x =& \frac{m-1}{m}\int \sin^{m-2}x \mathrm{d}x - \frac{1}{m}\sin^{m-1}x\cos x \end{aligned}

cosmxdx=cosm1xdsinx=sinxcosm1xsinxdcosm1x=sinxcosm1x+(m1)sin2xcosm2dx=sinxcosm1x+(m1)(1cos2x)cosm2dx=sinxcosm1x+(m1)cosm2dx(m1)cosmdxmcosmxdx=(m1)cosm2dx+sinxcosm1xcosmdx=m1mcosm2dx+1msinxcosm1x\begin{aligned} \int \cos^mx \mathrm{d}x &= \int \cos^{m-1}x\mathrm{d}\sin x \\ &= \sin x \cos^{m - 1}x - \int \sin x \mathrm{d}\cos^{m - 1}x\\ &= \sin x \cos^{m - 1}x + (m - 1) \int \sin^{2}x\cos^{m-2}\mathrm{d}x\\ &= \sin x \cos^{m - 1}x + (m - 1) \int (1-\cos^{2}x)\cos^{m-2}\mathrm{d}x\\ &= \sin x \cos^{m - 1}x + (m - 1) \int \cos^{m-2}\mathrm{d}x-(m-1)\int\cos^m\mathrm{d}x\\ m \int \cos ^m x\mathrm{d}x&=(m-1)\cos^{m-2}\mathrm{d}x + \sin x \cos^{m - 1}x\\ \int \cos^{m}\mathrm{d}x &= \frac{m - 1}{m}\int \cos^{m - 2}\mathrm{d}x + \frac{1}{m}\sin x \cos^{m - 1}x \end{aligned}

Im=tanmxdx=tanm2x(sec2x1)dx=tanm2xdtanxtanm2xdx=tanm1xtanxdtanm2xIm2=tanm1x(m2)tanm2(1+tan2x)dxIm2=tanm1x(m1)Im2(m2)ImIm=tanm1xm1Im2\begin{aligned} I_m=\int \tan^m x \mathrm{d}x &= \int \tan^{m - 2}x(\sec^{2}x - 1) \mathrm{d}x\\ &= \int \tan ^{ m - 2}x \mathrm{d}\tan x - \int \tan^{m - 2}x\mathrm{d}x\\ &= \tan^{m - 1}x - \int \tan x \mathrm{d} \tan^{m - 2}x - I_{m-2}\\ &= \tan^{m - 1}x- (m - 2)\int \tan^{m - 2}(1+\tan^{2}x)\mathrm{d}x - I_{m-2}\\ &= \tan^{m - 1}x - (m-1)I_{m-2} - (m - 2)I_m\\ I_m &= \frac{\tan^{m - 1}x}{m - 1} - I_{m - 2} \end{aligned}

一些比较重要的

好恶心啊。

  • dxa2x2\displaystyle \int \frac{\mathrm{d}x}{a^{2}-x^{2}}
  • dxa2+x2\displaystyle \int \frac{\mathrm{d}x}{a^{2}+x^{2}}
  • dxa2x2\displaystyle \int \frac{\mathrm{d}x}{\sqrt{a^{2}-x^{2}}},其中 a>0a>0
  • a2x2dx\displaystyle \int\sqrt{a^{2}-x^{2}}\mathrm{d}x,其中 a>0a>0
  • dxx2+a2\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}}
  • dxx2a2\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^2-a^{2}}},其中 a>0a>0
  • a2+x2dx\displaystyle \int\sqrt{a^2+x^2}\mathrm{d}x
  • x2a2dx\displaystyle \int\sqrt{x^{2}-a^{2}}\mathrm{d}x

对于第一个,我们发现可以使用拆项的方式,其即为 dxa2x2=12alnaxa+x+C\displaystyle \int \frac{\mathrm{d}x}{a^{2}-x^{2}}=\frac{1}{2a}\ln\left\vert \frac{a-x}{a+x} \right\vert +C

对于第二个,发现是 arctan\arctan 的形式,凑微分便有 dxa2+x2=1aarctanxa+C\displaystyle \int \frac{\mathrm{d}x}{a^{2}+x^{2}}=\frac{1}{a}\arctan \frac{x}{a}+C

对于第三个,发现是 arcsin\arcsin 的形式,凑微分便有 dxa2x2=arcsinxa+C\displaystyle \int \frac{\mathrm{d}x}{\sqrt{a^{2}-x^{2}}} = \arcsin \frac{x}{a} + C

对于 a2x2dx\displaystyle \int\sqrt{a^{2}-x^{2}}\mathrm{d}x,使用三角换元即可,不是特别恶心。令 x=asint(π2tπ2)x=a\sin t \displaystyle \left( -\frac{\pi}{2}\le t\le\frac{\pi}{2} \right),则

a2x2dx=a1sin2tacostdt=a2cos2dt=a22(1+cos2t)dt=a22(t+sintcost)+C=a22[arcsinxa+xa1(xa)2]+C=a22arcsinxa+x2a2x2+C\begin{aligned} \int\sqrt{a^{2}-x^{2}}\mathrm{d}x &= \int a\sqrt{1-\sin^{2}t}\cdot a\cos t \mathrm{d}t\\ &=a^2\int \cos^{2} \mathrm{d}t\\ &= \frac{a^{2}}{2}\int(1+\cos 2t)\mathrm{d}t\\ &= \frac{a^{2}}{2}\left( t+\sin t\cos t \right) +C\\ &=\frac{a^2}{2}\left[ \arcsin \frac{x}{a} + \frac{x}{a}\sqrt{1-\left( \frac{x}{a} \right) ^{2}} \right] +C\\ &= \frac{a^2}{2}\arcsin \frac{x}{a} + \frac{x}{2}\sqrt{a^{2}-x^{2}}+C \end{aligned}

对于 dxx2+a2\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}},同样可以使用三角换元,令 x=atant(π2<t<π2)\displaystyle x=a \tan t \left( -\frac{\pi}{2}< t < \frac{\pi}{2} \right),则

dxx2+a2=asec2tdta2sec2t=sectdt\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}} &= \int \frac{a \sec^{2}t \mathrm{d} t}{\sqrt{a^{2}\sec^{2} t}}= \int \sec t \mathrm{d}t\\ \end{aligned}

在继续之前,我们先处理 sec\sec 的积分,有 secxdx=cosxcos2xdx=d(sinx)1sin2x=12ln1+sinx1sinx+C\displaystyle \int \sec x \mathrm{d}x = \int \frac{\cos x}{\cos^{2}x}\mathrm{d}x = \int \frac{\mathrm{d}(\sin x)}{1-\sin^{2}x} = \frac{1}{2}\ln\left\vert \frac{1+\sin x}{1 - \sin x} \right\vert + C,事实上还可以继续化简,继续上下配凑:12ln1+sinx1sinx=12ln(1+sinx)2cos2x=lnsinx+1cosx=lntanx+secx\displaystyle \frac{1}{2}\ln\left\vert \frac{1+\sin x}{1-\sin x} \right\vert = \frac{1}{2}\ln\left\vert \frac{(1+\sin x)^2}{\cos^{2}x} \right\vert = \ln\left\vert \frac{\sin x + 1}{\cos x} \right\vert =\ln\left\vert \tan x + \sec x \right\vert,综上有 secxdx=lntanx+secx+C\displaystyle \int \sec x \mathrm{d}x = \ln \left\vert \tan x + \sec x \right\vert + C

回到上面,我们继续处理:

dxx2+a2=lntant+sect+C\int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}}=\ln\left\vert \tan t + \sec t \right\vert + C

在直角三角形中,可以看出来,当 tant=xa\displaystyle \tan t = \frac{x}{a} 时,sect=x2+a2a\displaystyle \sec t = \frac{\sqrt{x^{2}+a^{2}}}{a},故

dxx2+a2=lnxa+x2+a2a+C=lnx+x2+a2+C\int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}}=\ln\left\vert \frac{x}{a}+\frac{\sqrt{x^{2}+a^{2}}}{a} \right\vert +C = \ln\left\vert x+\sqrt{x^{2}+a^{2}} \right\vert +C

(最后一步将 lna\ln a 提出来扔进 CC 里面了),和上文的双曲换元比较,谁简单应该不用我说了。

接下来是 dxx2a2\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^{2}-a^{2}}},可以类似上文,当 x>ax > a 时令 x=asectx = a\sec t,然后一通操作,再分类讨论 x<ax < -a 的情况,好麻烦啊,本文略了。但是既然这让我们这么痛苦,为什么不考虑一下双曲换元呢?令 x>ax > ax=acosht(t>0)x = a\cosh t (t>0),则

dxx2a2=sinhtdtsinht=t=arcoshxa+C=ln(x+x2a2)+C\begin{aligned} \int \frac{\mathrm{d}x}{\sqrt{x^{2}-a^{2}}}&= \int \frac{\sinh t \mathrm{d}t}{\sinh t} = t = \operatorname{arcosh}\frac{x}{a} +C= \ln(x+\sqrt{x^{2}-a^{2}}) + C\\ \end{aligned}

对于 x<ax<-a 的讨论是类似的,上下的负号会被消去,最后都是一样的。所以 dxx2a2=lnx+x2a2+C\displaystyle \int \frac{\mathrm{d}x}{\sqrt{x^{2}-a^{2}}} = \ln \left\vert x+\sqrt{x^{2}-a^{2}} \right\vert +C

综上我们有很常见且重要的结论:

dxx2±a2=lnx+x2±a2+C\int \frac{\mathrm{d}x}{\sqrt{x^{2}\pm a^{2}}}=\ln\left\vert x+\sqrt{x^{2}\pm a^{2}} \right\vert +C

最后是 x2+a2dx\displaystyle \int\sqrt{x^{2}+a^{2}}\mathrm{d}xx2a2dx\displaystyle \int \sqrt{x^{2}-a^{2}}\mathrm{d}x,这两个既可以使用分部积分也可以使用双曲换元,这里两种思路都说一遍。

先看前者,我们直接分部积分莽的话,有

x2+a2dx=xx2+a2xdx2+a2=xx2+a2x2dxx2+a2=xx2+a2x2+a2dx+a2dxx2+a2=x2x2+a2+a22lnx+x2+a2+C\begin{aligned} \int\sqrt{x^{2}+a^{2}}\mathrm{d}x&=x\sqrt{x^{2}+a^{2}}-\int x\mathrm{d}\sqrt{x^{2}+a^{2}} \\ &= x\sqrt{x^{2}+a^{2}} - \int \frac{x^{2}\mathrm{d}x}{\sqrt{x^{2}+a^{2}}}\\ &= x\sqrt{x^{2}+a^{2}} - \int \sqrt{x^{2}+a^{2}}\mathrm{d}x + a^{2}\int \frac{\mathrm{d}x}{\sqrt{x^{2}+a^{2}}}\\ &= \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln\left\vert x+\sqrt{x^{2}+a^{2}} \right\vert +C \end{aligned}

或者我们换元,令 x=asinhtx = a\sinh t,则

x2+a2dx=a2cosh2tdt=a2cosh2t+12dt=a24cosh2td(2t)+a22t=a24sinh2t+a22t+C=a22sinhtcosht+a22t+C=a2x1+x2a2+a22lnxa+(xa)2+1+C=x2x2+a2+a22lnx+x2+a2+C\begin{aligned} \int\sqrt{x^{2}+a^{2}}\mathrm{d}x &= a^{2}\int \cosh^{2}t \mathrm{d}t \\ &= a^2\int \frac{\cosh 2t+1}{2}\mathrm{d}t\\ &= \frac{a^2}{4}\int \cosh 2t \mathrm{d}(2t) + \frac{a^2}{2}t\\ &= \frac{a^2}{4}\sinh 2t + \frac{a^{2}}{2}t+C\\ &= \frac{a^2}{2}\sinh t \cosh t + \frac{a^{2}}{2}t+C\\ &= \frac{a}{2}x\sqrt{1+\frac{x^{2}}{a^{2}}}+\frac{a^{2}}{2}\ln\left\vert \frac{x}{a}+\sqrt{\left( \frac{x}{a} \right) ^{2}+1} \right\vert +C\\ &= \frac{x}{2}\sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2}\ln\left\vert x+\sqrt{x^{2}+a^{2}} \right\vert +C \end{aligned}

再看后者;

x2a2dx=xx2a2xdx2a2=xx2a2x2x2a2dx=xx2a2x2a2dxa2dxx2a2=x2x2a2a22lnx+x2a2+C\begin{aligned} \int\sqrt{x^{2}-a^{2}} \mathrm{d}x &= x\sqrt{x^{2}-a^{2}}-\int x\mathrm{d}\sqrt{x^{2}-a^{2}}\\ &= x\sqrt{x^{2}-a^{2}}-\int \frac{x^{2}}{\sqrt{x^{2}-a^{2}}}\mathrm{d}x\\ &= x\sqrt{x^{2}-a^{2}}-\int \sqrt{x^{2}-a^{2}}\mathrm{d}x - a^{2}\int \frac{\mathrm{d}x}{\sqrt{x^{2} - a^{2}}}\\ &= \frac{x}{2}\sqrt{x^{2}-a^{2}}- \frac{a^{2}}{2}\ln\left\vert x+\sqrt{x^{2}-a^{2}} \right\vert +C\\ \end{aligned}

好像根本就差不多啊,然后一样,分讨,先考虑 x>ax>a,令 x=acoshtx = a\cosh t,则

x2a2dx=a2sinh2tdt=a2cosh2t12dt=a22t+a24sinh2t+C=a22t+a22sinhtcosht+C=a22ln(xa+(xa)21)+a2x(xa)21+C=x2x2a2a22ln(x+x2a2)+C\begin{aligned} \int \sqrt{x^{2} - a^{2}}\mathrm{d}x &= a^2\int \sinh^2 t \mathrm{d}t\\ &= a^{2}\int \frac{\cosh2t-1}{2}\mathrm{d}t\\ &= -\frac{a^{2}}{2}t + \frac{a^{2}}{4}\sinh 2t + C\\ &= -\frac{a^{2}}{2}t + \frac{a^{2}}{2}\sinh t \cosh t + C\\ &= -\frac{a^{2}}{2}\ln\left( \frac{x}{a}+\sqrt{\left( \frac{x}{a} \right) ^{2}-1} \right) +\frac{a}{2}x\sqrt{\left( \frac{x}{a} \right) ^{2}-1}+C\\ &= \frac{x}{2}\sqrt{x^{2}-a^{2}}-\frac{a^2}{2}\ln\left( x+\sqrt{x^{2}-a^{2}} \right) +C \end{aligned}

然后还要对于 x<ax<-a 讨论,此处略过,同样是负号没了,所以答案和上面一样的,注意绝对值的问题。

也可以总结一下,但是用的不太多:

x2±a2dx=x2x2±a2±a22lnx+x2±a2+C\int \sqrt{x^{2}\pm a^{2}}\mathrm{d}x = \frac{x}{2}\sqrt{x^{2}\pm a^{2}}\pm \frac{a^2}{2}\ln\left\vert x + \sqrt{x^{2}\pm a^{2}} \right\vert +C

还有一个特别恶心的递推:In=dt(t2+a2)n\displaystyle I_n=\int \frac{\mathrm{d}t}{\left( t^{2}+a^{2} \right) ^{n}},期中 nn 为大于 11 的整数,且 a>0a>0

定积分的一些内容

分部积分的一个递推:考虑我们要求 In=0π2sinnxdx\displaystyle I_n = \int_0^{\frac{\pi}{2}}\sin^nx \mathrm{d}x,其中 nn 为大于 11 的整数,则:

In=0π2sinn1xdcosx=0π2cosxdsinn1xsinn1xcosx0π2=(n1)0π2sinn2xcos2xdx=(n1)0π2(1sin2x)sinn2xdx=(n1)In2(n1)InIn=n1nIn2\begin{aligned} I_n &= -\int_0^{\frac{\pi}{2}}\sin^{n-1}x\mathrm{d}\cos x \\ &= \int_0^{\frac{\pi}{2}}\cos x \mathrm{d}\sin^{n-1}x - \sin^{n-1}x \cos x\big|_0^{\frac{\pi}{2}}\\ &= (n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}x \cos^{2}x\mathrm{d}x\\ &= (n-1)\int_0^{\frac{\pi}{2}}(1-\sin^{2}x)\sin^{n-2}x\mathrm{d}x\\ &=(n-1)I_{n-2}-(n-1)I_n\\ I_n&= \frac{n-1}{n}I_{n-2} \end{aligned}

I0=π2I_0 = \displaystyle \frac{\pi}{2}I1=1I_1=1,所以把递推式展开,有

0π2sin2kxdx=(2k1)!!(2k)!!π20π2sin2k+1xdx=(2k)!!(2k+1)!!\int_0^{\frac{\pi}{2}}\sin^{2k}x\mathrm{d}x=\frac{(2k-1)!!}{(2k)!!}\cdot \frac{\pi}{2}\\ \int_0^{\frac{\pi}{2}}\sin^{2k+1}x\mathrm{d}x=\frac{(2k)!!}{(2k+1)!!}

其中 k=1,2,k=1,2, \cdots

并且由于若我们令 t=π2xt = \displaystyle \frac{\pi}{2}-x,则

0π2cosnxdx=π20sinntd(t)=0π2sinnxdx\int_0^{\frac{\pi}{2}}\cos^nx\mathrm{d}x=\int _{\frac{\pi}{2}}^0\sin^n t \mathrm{d}(-t) = \int_0^{\frac{\pi}{2}}\sin^nx\mathrm{d}x

所以上面的递推式的形式对 cos\cos 完全适用。


一些高数积分笔记
https://blog.imyangty.com/note-integral/
Author
YangTY
Posted on
November 3, 2023
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