what is the integral of sin^(-1)x dx?

Answer 1

arcsin(x) dx
Let u = arcsin(x) and dv = dx, then
du = 1/sqrt(1-x^2) and v = x
uv - v du
= x arcsin(x) - integral of [x/sqrt(1-x^2) dx]
Use substitution. Let u = 1-x^2, then du = -2x dx or (-1/2)du = x dx
Answer: x arcsin(x) + sqrt(1-x^2) + C

Answer 2

xsin-1(x) + (1-x^2)^1/2+C

use integrating by parts

by the way, sin^-1(x) does not = 1/sinx and does not =cscx

Answer 3

sin^(-1)x = csc (x)

the integral of csc(x) is
ln | tan (x/2) | + C