Find the solution to y′=cos(x)sec2(y)y' = \cos(x) \sec^2(y)y′=cos(x)sec2(y) with y(0)=0y(0) = 0y(0)=0.
tan(y)=sin(x)\tan(y) = \sin(x)tan(y)=sin(x)
13tan3(y)=sin(x)\frac{1}{3}\tan^3(y) = \sin(x)31tan3(y)=sin(x)
tan(y)=cos(x)\tan(y) = \cos(x)tan(y)=cos(x)
sin(y)=tan(x)\sin(y) = \tan(x)sin(y)=tan(x)