Find the value of: ∫2cosx(1−sinx)(1+sin2x)dx

Consider, I = ∫2cosx(1−sinx)(1+sin2x)dx. Substituting sinx = u ⇒ du = cosxdx, we get, I = ∫2cosxdx(1−sinx)(1+sin2x) = ∫2(1−u)(1+u²)du I = −∫(u−1−u³+u²)du I = −[u²/2 - u - u⁴/4 + u³/3] + C Substituting back, we get: I = −[sin²x/2 - sinx - sin⁴x/4 + sin³x/3] + C where C is a real constant of integration.