Question

∫sin^-1x dx/√ 1-x^2 = ?

Answer 1

did you try to put
\(\large x = \sin t\)
?

Answer 2

when \(x =\sin t \\ \sin^{-1}x = t \\ dx =....?\)

Answer 3

I tried to do it like this
Let,1-x^2=z
=>d/dx(1)-d/dx x^2=dz/dx

Answer 4

good try, but you'll get stuck after some steps.

can you please try the substitution,
x= sin t now ?

Answer 5

Ok thanks

Answer 6

tell me if you get stuck with that or couldn't find the correct final answer.

Answer 7

sure i will..let me try it..and i will tell u the answer..if u can check out please after that .

Answer 8

sure :)

Answer 9

let,sin^-1x=z
or,d/dx sin^-1x=dz/dx
or,1/√ 1-x^2=dz/dx
or,dx/√ 1-x^2=dz

Now,∫sin^-1x dx/√ 1-x^2=∫z dz=z^1+1/1+1+c=z^2/2+c= (sin^-1x)^2/2+c is this the answer @hartnn

Answer 10

you're absolutely correct!
good work :)

Answer 11

thank you :)

Answer 12

most welcome ^_^

Answer 13

^_^