Is it possible to integrate arccos(x)?

Question

Or any inverse trig function like that?

Answer 1

Integration [arccos(x)]

u = arc cosx
v = 1
du/dx = -1/sqrt(1 - x^2)
Integration (vdx) = x

Integration [arccos(x)]
= (arc cosx) (x) + Integration [x / sqrt(1 - x^2)]

Substitute (1 - x^2) = u
Differentiating
-2xdx = du

Integration [x / sqrt(1 - x^2)]
= Integration[- 1/2 . 1/sqrt(u) . du]
= - sqrt(u)
= - sqrt(1 - x^2)

Integration [arccos(x)]
= x arccosx - sqrt(1 - x^2) + c >> c = constant of integ...

Answer 2

integral arccos(x) dx = xarccos(x) - sqrt(1-x^2) + C
integral arcsin(x) dx = xarcsin(x) +sqrt(1-x^2) + C
integral arctan(x) dx = xarctan(x) -.5* ln(1+x^2) + C