integral
Posted: 04/04-2007 19:07
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Posted: 04/04-2007 19:47
Kaller integralet ditt for I:
[tex]I\,=\,sqrt{3}\int_0^1{1\over \cos^2(\frac{\pi}{6}(x+1))}{\rm dx}[/tex]
[tex]I\,=\, sqrt3 \cdot\,\frac{6}{\pi}[ \tan(\frac {\pi}{6}(x+1))]_0^1\,=\,\frac{6sqrt3}{\pi}(sqrt3\,-\,\frac{1}{sqrt3})\,=\,\frac{12}{\pi}[/tex]
[tex]I\,=\,sqrt{3}\int_0^1{1\over \cos^2(\frac{\pi}{6}(x+1))}{\rm dx}[/tex]
[tex]I\,=\, sqrt3 \cdot\,\frac{6}{\pi}[ \tan(\frac {\pi}{6}(x+1))]_0^1\,=\,\frac{6sqrt3}{\pi}(sqrt3\,-\,\frac{1}{sqrt3})\,=\,\frac{12}{\pi}[/tex]
Posted: 04/04-2007 19:51
å ja, det glemte jeg, takk!