Question 461859
Presuming you mean


*[tex \LARGE \int \sqrt{1 - y^2}\,dy]


Then you should make the substitution y = sin u. Then, dy = cos u du and we can substitute y and dy:


*[tex \LARGE \int \sqrt{1 - \sin^2{u}} \cos{u}\, du = \int \cos^2 {u}\, du] by the Pythagorean identity. Applying the power reduction identity,


*[tex \LARGE \int \cos^2 {u}\, du = \int \frac{1}{2} + \frac{1}{2} \cos{2u}\, 
du]


*[tex \LARGE = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C]


Since u is the inverse sine of y,


*[tex \LARGE \frac{1}{2}(sin^{-1}{y}) + \frac{1}{4}\sin{(2\sin^{-1}{y})} + C]


I'll let you simplify the second term (use some trig for this!).