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Integral

Posted: 24/02-2007 17:55
by al-Khwarizmi
Noen løsningsforslag til denne?

-0,5 [symbol:integral] x^2/ [symbol:rot] (1-x^2) dx ??

Posted: 24/02-2007 18:28
by ingentingg
[tex]x = \sin u\\dx = \cos u du\\ -\frac12\int\frac{x^2}{\sqrt{1-x^2}}dx = -\frac12\int\frac{\sin^2u}{\sqrt{1-\sin^2u}}\cos u du = -\frac12\int\frac{\sin^2u}{\sqrt{\cos^2u}}\cos u du = -\frac12\int sin^2u[/tex]