Question

integration...

Answer 1

integrate x^2 sinx

Answer 2

This is a fairly lengthy problem to work out from scratch. You will have to apply integration by parts twice.

Answer 3

i just need the answer

Answer 4

I won't just give you the answer. First, we need to identify the \(u,v\) for the first time. What do you think they should be?

Answer 5

i got -x^2 cosx + 2xsinx + 2cosx + C

Answer 6

wolfram puts it in a different form and isn't clear to me

Answer 7

Well, let's do it by hand then. What do you think the \(u,v\) should be?

Answer 8

yes

Answer 9

i used tabular method

Answer 10

so u was x^2... dv=sinx

Answer 11

Alright. Then \(du=2x\) and \(v=-\cos(x)\). So after one step of integration, we get\[-x^2\cos(x)-\int-\cos(x)2x dx=-x^2\cos(x)+2\int x\cos(x)dx\]Now we integrate by parts again.

Answer 12

This time with \(u=x\), and \(dv=\cos(x)\). We get\[-x^2\cos(x)+2\left(x\sin(x)-\int \sin(x)dx\right)\]Still following?

Answer 13

yes

Answer 14

Now we do that last integral and distribute the 2 to get \[-x^2\cos(x)+2x\sin(x)+2\cos(x)+C\]

Answer 15

ok, that's what i got!

Answer 16

Then you're right.

Answer 17

i couldn't tell that from just looking at wolf

Answer 18

but anyways, thank you!

Answer 19

You're welcome.