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appleduardo
whats the integral of arc sin x ??
so far i have:\[x(arcsinx)-\int\limits_{}^{}x \frac{ 1 }{ \sqrt{1-x^2} }dx\]
Looks like you've done most of the hard part. Now take 1-x^2 and integrate by subtitution.
uhmm how can i do that?
there's where i got confused!
trig substitution x = sin(u) dx = cos(u) du
So take (1-x^2)=u and take the derivative with respect to x, this becomes -2x=du/dx which can be rearranged with algebra into: xdx=-(1/2)du Substitution back into the original integrand gives: \[-\int\limits_{}^{}\frac{- du }{ 2\sqrt{u} } = \frac{ 1 }{ 2 }\int\limits_{}^{}\ u^{-1/2} du\]
@dumbcow trig substitution isn't really useful here.