calculus
posted by waqas .
integrate it by integrating factor
(cos^3x)dy/dx +ycosx=sinx

Hmmm. rearrange things a bit to get
y' + sec^2 x y = tan x sec^2 x
that is
y' + P(x)y = Q(x)
integrating factor is thus e^∫sec^2 x dx = e^(tan x)
now plug and chug