# Homework Help: More integration by parts

1. Feb 9, 2012

### bobsmith76

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.

Last edited: Feb 9, 2012
2. Feb 9, 2012

### ehild

You missed a pair of parentheses.

ehild

3. Feb 9, 2012

### Sourabh N

There's your answer for the first part of your question. For the second part, remember that ∫ Sin[x] = -Cos[x]