Question

what is the intergral of arc sin2X

Answer 1

integral sin^(-1)(2 x) dx
For the integrand sin^(-1)(2 x), substitute u = 2 x and du = 2 dx:
= 1/2 integral sin^(-1)(u) du
For the integrand sin^(-1)(u), integrate by parts, integral f dg = f g- integral g df, where
f = sin^(-1)(u), dg = du,
df = 1/sqrt(1-u^2) du, g = u:
= 1/2 u sin^(-1)(u)-1/2 integral u/sqrt(1-u^2) du
For the integrand u/sqrt(1-u^2), substitute s = 1-u^2 and ds = -2 u du:
= 1/4 integral 1/sqrt(s) ds+1/2 u sin^(-1)(u)
The integral of 1/sqrt(s) is 2 sqrt(s):
= sqrt(s)/2+1/2 u sin^(-1)(u)+constant
Substitute back for s = 1-u^2:
= sqrt(1-u^2)/2+1/2 u sin^(-1)(u)+constant
Substitute back for u = 2 x:
= 1/2 sqrt(1-4 x^2)+x sin^(-1)(2 x)+constant