Question

evaluate the integral: S arccos(x)dx

Answer 1

Integration by parts:

integral(uv) = uv - integral(vdu)

u=arccos(x)
du = - 1 / √(1 - x^2)
dv = dx
v = x

Therefore: uv = xarccos(x)

integral(vdu) = x/√(1-x^2)

let k = (1-x²)
then dk = -2x dx
and -dk/2 = x dx

∫x/(1-x²) * dx = (-1/2)∫dk/√k = -√k

Therefore: -√k = -√(1-x^2)

Final answer: integral(uv) = uv - integral(vdu)

= xarccos(x) -√(1-x^2) +C

Answer 2

hope it helped !