How Do You Solve ∫ x^2 sin x Using Integration by Parts?

[bobsmith76]

Homework Statement

∫ x² sin x

Homework Equations

uv - ∫ v du

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.

 

[ehild]

Homework Statement

∫ x² sin x

Homework Equations

uv - ∫ v du

The Attempt at a Solution

u = x²

du = 2x

dv = sin x

v = -cos x

step 1. x² - cos x - ∫ -cos x 2x

You missed a pair of parentheses.

ehild

 

[Sourabh N]

You missed a pair of parentheses.

ehild

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