Integration by sustitution

[Firescape]

I can't integrate dx/(1-x^2)^1/2 inverse sinx

[Cap'n Refsmmat]

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.

[Firescape]

no, i mean dx/[(1-x^2)^1/2.inverse sin x]

[Cap'n Refsmmat]

So [math]\frac{dx}{(1-x^2)^{1/2} \sin^{-1} x}[/math]?

[Firescape]

Correct! Thanks

[K!!]

[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].