Indefinite integral: SiNx ^ 3cosx ^ 2

Indefinite integral: SiNx ^ 3cosx ^ 2

∫sinx^3cosx^2dx
=-∫sin^2xcos^2xdcosx
=-∫(1-cos^2x)cos^2xdcosx
=-∫(-cos^4x+cos^2x)dcosx
=(1/5)cos^5x-(1/3)cos^3x+C

Indefinite integral of SiNx ^ 3cosx ^ 2

∫sinx^3cosx^2dx
=-∫sin^2xcos^2xdcosx
=-∫(1-cos^2x)cos^2xdcosx
=-∫(-cos^4x+cos^2x)dcosx
=(1/5)cos^5x-(1/3)cos^3x+C

Solving indefinite integral: ∫ (SiNx) ^ 2DX

∫ Sin & sup2; xdx = 1 / 2 ∫ (1-cos2x) DX = 1 / 2 (x - ∫ cos2xdx) = 1 / 2 (x-1 / 2 ∫ cos2xd2x) = 1 / 2 (x-1 / 2sin2x) = x / 2 - (sin2x) / 4 checking calculation: X / 2 - (sin2x) / 4 '= 1 / 2-2 / 4 cos2x = 1 / 2-1 / 2 (COS & sup2; x-sin & sup2; x) = 1 / 2 (1-cos & sup2; X + Sin & sup2; x) = 1 /