What is the antiderivative of arccosine?

Answer 1

Well obviously you can see that if,
arccos x = y
you get
x = cos y
dx/dy = -sin y = -V(1-x^2)
so dy/dx = d(arccos x) /dx = -1/V(1-x^2)
where V is the square root radical.
So use integration by parts for int. arccos x dx
with u=arccos x and v=1
that is in the formula you differentiate u and integrate 1, then,
int. arccos x dx = x.arccos x + int. x/V(1-x^2) dx
for the integral on the right hand side use a substitution like t=V(1-x^2)
Et voila the integral has been evaluated!
Hope this helps!

Answer 2

Substitute x=cosy. Then dx = -siny dy, and arccosx dx becomes
-y*siny dy.

Integrate by parts: u = y, dv = -siny dy, du = dy, v = cosy
integral (udv) = uv - integral(vdu) =
y*cosy - integral(cosy dy)
= y*cosy - siny + C
= arccosx * x - sqrt(1-cos²y) + C
= x*arccosx - sqrt(1-x²) + C