Question Detail
Question
Obtain ∫x²cosx dx
Correct Answer
Option B is the correct answer.
Detailed Explanation
This integral requires integration by parts, ∫ u dv = uv - ∫ v du, applied twice.First application:Let u = x², dv = cosx dxThen du = 2x dx, v = sinx∫x²cosx dx = x²sinx - ∫sinx (2x) dx = x²sinx - 2∫xsinx dxSecond application (for ∫xsinx dx):Let u = x, dv = sinx dxThen du = dx, v = -cosx∫xsinx dx = x(-cosx) - ∫(-cosx) dx= -xcosx + ∫cosx dx= -xcosx + sinx + C'Substitute the result of the second integral back into the first expression:∫x²cosx dx = x²sinx - 2(-xcosx + sinx) + C= x²sinx + 2xcosx - 2sinx + C
Hint
Apply integration by parts (∫ u dv = uv - ∫ v du) twice. For the first application, let u = x² to reduce the power of x. For the second, let u = x.