Question

Integral Question (posted in comments)

Answer 1

\[\int\limits_{}^{} \frac{ \sin^{-1} x }{ \sqrt{1-x^2} }dx\]

Answer 2

let y= sin^-1 x then dy/dx = dx/ sqrt(1-x^2)
now substitute

Answer 3

integrate in terms of y after that just again substitute the value of y

Answer 4

okay so the answer is \[\frac{ \ (sin^{-1} x) ^2}{ 2 }+C\] but i got \[\sin^{-1}( \frac{ x }{ 1 })+C\]do you think you (or anyone else) know where i went wrong?

Answer 5

and can you tell me how you got there?

Answer 6

Well my notes say once you arrive at du/sqrt(a^2-u^2) that you just put in the respective numbers and get sin^-1(u/a)+C.

Answer 7

i realized my mistake!
thank you!

Answer 8

good work (: