1. The problem statement, all variables and given/known data ∫ x2 sin x 2. Relevant equations uv - ∫ v du 3. The attempt at a solution u = x2 du = 2x dv = sin x v = -cos x step 1. x2 - cos x - ∫ -cos x 2x I think -cos x * 2x becomes -2x cos x so now we have step 2. x2 - cos x - ∫ -2x cos x which means I have to integrate by parts again. Here, concentrating on just the right hand side u = 2x du = 2 dv = cos x v = sin x step 3. 2x sin - ∫ sin x * 2 [after 10 minutes of research I've decided that I have to move that 2 to the left of the integral. That sort of helps. previously I took the antiderivative of 2.] step 4. 2x sin + 2 -cosx now add the left hand side part from above step 5. x2 - cos x step 6. x2 - cos x + 2x sin x + - 2 cosx + C the book says the answer is -x2 cos x + 2x sin x + 2 cos x + C So I'm almost correct, I just don't understand how they got the negative on x2, Also my right cos x is negative and their's is positive.