Løs integralet:
[tex]\int \frac{1}{a^2+cos^2(x)}dx[/tex]
ved hjelp av substitusjon [tex]\: u=tan (x) \:[/tex]
Hvordan?
integral
Moderatorer: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
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Integralen
- von Neumann
- Innlegg: 525
- Registrert: 03/10-2010 00:32
Sist redigert av Integralen den 12/01-2012 23:31, redigert 1 gang totalt.
-
Integralen
- von Neumann
- Innlegg: 525
- Registrert: 03/10-2010 00:32
[tex]u=tan(x)[/tex]
[tex]du=\frac{1}{cos^2(x)}dx[/tex]
hvordan skrive [tex]\: cos^2(x)\:[/tex]
uttrykt ved u?
[tex]du=\frac{1}{cos^2(x)}dx[/tex]
hvordan skrive [tex]\: cos^2(x)\:[/tex]
uttrykt ved u?
-
Integralen
- von Neumann
- Innlegg: 525
- Registrert: 03/10-2010 00:32
[tex]tan(x)=\frac{sin(x)}{cos(x)}[/tex]
[tex]cos^2(x)=\frac{sin^2(x)}{tan^2(x)}[/tex]
[tex]sin^2(x)=1-cos^2(x)[/tex]
[tex]cos^2(x)+\frac{cos^2(x)}{tan^2(x)}=\frac{1}{tan^2(x)}[/tex]
[tex]cos^2(x)=\frac{1}{tan^2(x)+1}[/tex]
[tex]cos^2(x)=\frac{1}{1+u^2}[/tex]
[tex]\int \frac{1}{a^2 +cos^2(x)}dx[/tex]
[tex]u=tan(x)[/tex]
[tex]du=\frac{1}{cos^2(x)}dx[/tex]
[tex]\frac{1}{a^2} \int \frac{1}{(\frac{1}{\frac{1}{1+u^2}}+\frac{1}{a^2})} du=\frac{1}{(a^2+1)} \int \frac{1}{1+(\frac{au}{\sqrt{a^2+1}})^2}du=\frac{\arctan(\frac{atan(x)}{\sqrt{a^2+1}})}{a\sqrt{a^2+1}}+C[/tex]
q.e.d
[tex]cos^2(x)=\frac{sin^2(x)}{tan^2(x)}[/tex]
[tex]sin^2(x)=1-cos^2(x)[/tex]
[tex]cos^2(x)+\frac{cos^2(x)}{tan^2(x)}=\frac{1}{tan^2(x)}[/tex]
[tex]cos^2(x)=\frac{1}{tan^2(x)+1}[/tex]
[tex]cos^2(x)=\frac{1}{1+u^2}[/tex]
[tex]\int \frac{1}{a^2 +cos^2(x)}dx[/tex]
[tex]u=tan(x)[/tex]
[tex]du=\frac{1}{cos^2(x)}dx[/tex]
[tex]\frac{1}{a^2} \int \frac{1}{(\frac{1}{\frac{1}{1+u^2}}+\frac{1}{a^2})} du=\frac{1}{(a^2+1)} \int \frac{1}{1+(\frac{au}{\sqrt{a^2+1}})^2}du=\frac{\arctan(\frac{atan(x)}{\sqrt{a^2+1}})}{a\sqrt{a^2+1}}+C[/tex]
q.e.d