Question

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

Answer 1

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

Answer 2

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))

Answer 3

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

Answer 4

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

Answer 5

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

*typo no 2 in front of the sqrt

Answer 6

sorry thats a '+'

Answer 7

thanks Zed. Thanks everyone for helping!

Answer 8

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

Answer 9

yw

Answer 10

you got there with integration by parts or substitution?

Answer 11

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

Answer 12

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