for the integral of sin^(-1)xdx I get
xsin^(-1) - (1-x^2)^(1/2) + C
....would this be correct?

Question

for the integral of sin^(-1)xdx  I get
xsin^(-1) - (1-x^2)^(1/2) + C
....would this be correct?

anonymous · Answer

integral of sin^-1xdx= xsin^-1(x) + 2sqrt(1-x^2) + C

anonymous · Answer

Is that one of those integrals that I just need to memorize or did you do some sort u-substitution. I thought the derivative of $$\sin^{-1} x    $$ was 1/(sqrt(1-x^2))

anonymous · Answer

sqrt(1-x^2) - x sin(-1)x + c

anonymous · Answer

oh sorry, i was wrong.. i am very much sorry...

anonymous · Answer

integral = xsin^-1(x) + sqrt(1-x^2) + C

*typo no 2 in front of the sqrt

anonymous · Answer

sorry thats a '+'

anonymous · Answer

thanks Zed. Thanks everyone for helping!

anonymous · Answer

sqrt(1-x^2) + x sin(-1)x + c

anonymous · Answer

yw

anonymous · Answer

you got there with integration by parts or substitution?

anonymous · Answer

integral udv=uv- integral vdu, 
just use this theorem

anonymous · Answer

first integration by parts then for round two i did u substitution