Lys van integrale

Integrasie is een van die hoofbewerkings van kalkulus. Vir differensiasie kan die eenvoudiger dele van 'n funksie maklik gedifferensieer word, wat differensiasie dan vergemaklik, maar dit kan egter nie met integrasie gedoen word nie. Vir gevalle waar daar met komplekse funksies gewerk word, is dit makliker om 'n lys van integrale byderhand te hou.

Vir die doeleindes van hierdie lys word K as arbitrêre-integrasiekonstante gebruik.

Reëls by die integreer van algemene funksies

$\int af(x)dx=a\int f(x)dx\text{(}a\text{ konstant)}$

$\int [f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx$

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int [f^{\prime }(x)(\int g(x)dx)]dx$

$\int [f(x){]}^n{f}^{\prime }(x)dx=\frac{[f(x){]}^{n+1}}{n+1}+K\text{(vir }n\ne -1\text{)}$

$\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+K$

$\int f^{\prime }(x)f(x)dx=\frac{1}{2}[f(x){]}^2+K$

Integrale van eenvoudige funksies

Rasionale funksies

$\int \mathrm{d}x=x+K$

$\int x^n\mathrm{d}x=\frac{x^{n+1}}{n+1}+K\text{ mits }n\ne -1$

$\int \frac{dx}{x}=\ln \left|x\right|+K$

$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\text{(bgtan)}\frac{x}{a}+K$

Irrasionale funksies

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+K$

$\int \frac{-dx}{\sqrt{a^2-x^2}}={\cos }^{-1}\frac{x}{a}+K$

$\int \frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}{\sec }^{-1}\frac{|x|}{a}+K$

Logaritmes

$\int \ln xdx=x\ln x-x+K$

$\int {\log }_{b}xdx=x{\log }_{b}x-x{\log }_{b}e+K$

Eksponensiaalfunksies

$\int e^{x}dx=e^x+K$

$\int a^{x}dx=\frac{a^x}{\ln a}+K$

Trigonometriese funksies

$\int \sin xdx=-\cos x+K$

$\int \cos xdx=\sin x+K$

$\int \tan xdx=\ln |\sec x|+K$

$\int \cot xdx=-\ln |\csc x|+K$

$\int \sec xdx=\ln |\sec x+\tan x|+K$

$\int \csc xdx=-\ln |\csc x+\cot x|+K$

$\int {\sec }^{2}xdx=\tan x+K$

$\int {\csc }^{2}xdx=-\cot x+K$

$\int \sec x\tan xdx=\sec x+K$

$\int \csc x\cot xdx=-\csc x+K$

$\int {\sin }^{2}xdx=\frac{1}{2}(x-\sin x\cos x)+K$

$\int {\cos }^{2}xdx=\frac{1}{2}(x+\sin x\cos x)+K$

$\int {\sec }^{3}xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln |\sec x+\tan x|+K$

$\int {\sin }^{n}xdx=-\frac{{\sin }^{n-1}x\cos x}{n}+\frac{n-1}{n}\int {\sin }^{n-2}xdx$

$\int {\cos }^{n}xdx=\frac{{\cos }^{n-1}x\sin x}{n}+\frac{n-1}{n}\int {\cos }^{n-2}xdx$

$\int \text{bgtan}xdx=x\text{bgtan}x-\frac{1}{2}\ln |1+x^2|+K$

Hiperboliese funksies

$\int \sinh xdx=\cosh x+K$

$\int \cosh xdx=\sinh x+K$

$\int \tanh xdx=\ln |\cosh x|+K$

$\int \text{csch}xdx=\ln |\tanh \frac{x}{2}|+K$

$\int \text{sech}xdx=\text{bgtan}(\sinh x)+K$

$\int \coth xdx=\ln |\sinh x|+K$

$\int {\text{sech}}^{2}xdx=\tanh x+K$

Inverse hiperboliese funksies

$\int {\sinh }^{-1}xdx=x{\sinh }^{-1}x-\sqrt{x^2+1}+K$

$\int {\cosh }^{-1}xdx=x{\cosh }^{-1}x-\sqrt{x^2-1}+K$

$\int {\tanh }^{-1}xdx=x{\tanh }^{-1}x+\frac{1}{2}\log (1-x^2)+K$

$\int {\text{csch}}^{-1}xdx=x{\text{csch}}^{-1}x+\log \left[x(\sqrt{1+\frac{1}{x^2}}+1)\right]+K$

$\int {\text{sech}}^{-1}xdx=x{\text{sech}}^{-1}x-\text{bgtan}\left(\frac{x}{x-1}\sqrt{\frac{1-x}{1+x}}\right)+K$

$\int {\coth }^{-1}xdx=x{\coth }^{-1}x+\frac{1}{2}\log (x^2-1)+K$

Bepaalde integrale sonder geslote-vorm afgeleides

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sqrt{x}{e}^{-x}dx=\frac{1}{2}\sqrt{\pi }$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\frac{1}{2}\sqrt{\pi }$ (die Gaussiese integraal)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x-1}dx=\frac{{\pi }^2}{6}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^3}{e^x-1}dx=\frac{{\pi }^4}{15}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (x)}{x}dx=\frac{\pi }{2}$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n}xdx=\frac{1\cdot 3\cdot 5\cdot \cdots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot n}\frac{\pi }{2}$ (mits n 'n ewe heelgetal en $n\ge 2$)

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n}xdx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{n}xdx=\frac{2\cdot 4\cdot 6\cdot \cdots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \cdots \cdot n}$ (mits n 'n onewe heelgetal en $n\ge 3$)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2}x}{x^2}dx=\frac{\pi }{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{z-1}{e}^{-x}dx=\mathrm{\Gamma }(z)$ (waar $\mathrm{\Gamma }(z)$ die Gamma funksie is)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\sqrt{\frac{\pi }{a}}\exp \left[\frac{b^2-4ac}{4a}\right]$ (waar $\exp [u]$ die eksponensiaalfunksie $e^u$ is.)

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta }d\theta =2\pi I_0(x)$ (waar $I_0(x)$ die gewysigde Bessel funksie van die eerste tipe is.)

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\sqrt{x^2+y^2}$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(1+x^2/\nu {)}^{-(\nu +1)/2}dx=\frac{\sqrt{\nu \pi }\mathrm{\Gamma }(\nu /2)}{\mathrm{\Gamma }((\nu +1)/2))}$ ($\nu >0$.

$\underset{a}{\overset{b}{\int }}f(x)dx=(b-a)\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{2^n-1}{\left(-1\right)}^{m+1}{2}^{-n}f(a+m\left(b-a\right)2^{-n})$

Verwysings

1. Stewart, J. (2003). Single Variable calculus. (5th ed.). Belmont, USA: Thomson Learning.
2. Groenewald, G.J., Hitge, M. (2005). Analise II Studiegids vir WISK121A. Potchefstroom: Noordwes-Universiteit.
3. Jordan, D.W., Smith, P. (2002). Mathematical techniques: An introduction for the engineering, physical and mathematical sciences. USA: Oxford University Press.

Aantekeninge

• Sien ook Lys van afgeleides
• Hierdie artikel is 'n vertaling van die Engelse Wikipedia artikel Table of integrals