## anonymous 5 years ago integral dx over square root of quantity of 2x-x^2

1. anonymous

Focus on the quantity within the square root You can rewrite 2x -x^2 as 1- (x^2 -2x +1) this gives you $\int\limits_{}^{}dx / (\sqrt{1-(x-1)^{2}}$ now put x-1=sin t dx=cost dt the above expression converts to costdt/[cost] =$\int\limits_{}^{}$dt =t =$\sin^{-1} (x-1)$

2. anonymous

you are my hero!!!