Hello,

$\displaystyle \int_{ 0}^{1/2} {cos^{-1}(x)dx}$

let

$\displaystyle u = cos^{-1}x$ and $\displaystyle dv=dx$

$\displaystyle du = \frac{-1}{\sqrt{1-x^2}}dx$ and $\displaystyle v=x$

Then:

$\displaystyle = [xarccos(x)]_{0}^{1/2} - \int_{0}^{1/2} \frac{-x}{\sqrt{1-x^2}}dx$

However at this point I would use u-sub to find the value of the second integral, but the answer has changed the limits of integration? I'm confused. Any help would be appreciated.