How does integrating arcsin(w) give sqrt(1-w²) + C?

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3ephemeralwnd
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can someone explain what happened after the part where it says "= warcsinw - integral of w/sqrt(1-w^2) = ... " ? i don't see how they got sqrt(1-w^2) + C
 
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  • Integration by parts of arcsin w: u = arcsin w, v' = 1, giving u' = 1/sqrt(1-w^2) and v = w; the integral of arcsin w dw equals w arcsin w minus the integral of w/sqrt(1-w^2) dw, which equals w arcsin w + sqrt(1-w^2) + C.
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3ephemeralwnd said:
can someone explain what happened after the part where it says "= warcsinw - integral of w/sqrt(1-w^2) = ... " ? i don't see how they got sqrt(1-w^2) + C
Ordinary substitution, with u = 1 - w2, and du = -2wdw.