Also i'm not positive if its asking for the integral of arccos or cos(x)^-1. I have the calculus ed.5 stewart. we are using it in math 125 at the university of Washington if that helps??
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What is the antiderivative (cos^-1(x) dx) from 0 to 1/2?
Verified answer
use integ by parts
u=arccosx
du=-1/sqrt(1-x^2)dx
dv=dx
v=x
xarccox+intx/sqrt(1-x^2)dx
to integrate the second part
let 1-x^2=k
-2xdx=dk
-1/2int1/sqrtkdk=-sqrtk
so the whole thingis xarccosx-sqrt(1-x^2)
the value is 1/2*pi/3-sqrt3/2+1=pi/6-sqrt3/2+1
=(pi+6-3sqrt3)/6
actually i think its asking for the integral of cos inverse (x) which is represented by cos^-1(x)
The only way to do this is by using by parts formula , assuming 1 as the second function and cos^-1(x) as the function.
the integral becomes (after applying by parts becomes )
cos^-1(x) * x + integral(x/sqrt(1-x^2))
which i think you can integrate easily