Firescape Posted December 1, 2007 Share Posted December 1, 2007 I can't integrate dx/(1-x^2)^1/2 inverse sinx Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted December 1, 2007 Share Posted December 1, 2007 Do you mean [math]\left ( \frac{dx}{1 - x^2} \right ) ^{\frac{1}{2}}\sin^{-1} x[/math]? It's hard to tell in text. Link to comment Share on other sites More sharing options...
Firescape Posted December 1, 2007 Author Share Posted December 1, 2007 no, i mean dx/[(1-x^2)^1/2.inverse sin x] Link to comment Share on other sites More sharing options...
Cap'n Refsmmat Posted December 1, 2007 Share Posted December 1, 2007 So [math]\frac{dx}{(1-x^2)^{1/2} \sin^{-1} x}[/math]? Link to comment Share on other sites More sharing options...
Firescape Posted December 1, 2007 Author Share Posted December 1, 2007 Correct! Thanks Link to comment Share on other sites More sharing options...
K!! Posted January 2, 2008 Share Posted January 2, 2008 [math]\int {\frac{{dx}} {{\sqrt {1 - x^2 } \cdot \arcsin x}}} = \int {\frac{{(\arcsin x)'}} {{\arcsin x}}\,dx} = \ln \left| {\arcsin x} \right| + k[/math]. Link to comment Share on other sites More sharing options...
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