# Thread: Integration by parts, basic,.. I think!

1. ## Integration by parts, basic,.. I think!

Use integration by parts to solve the following integrals:

(I don't know how to do an integral sign on this lol)

[integral] x^2 sin(3x) dx.

Now, I've solved it out and come to the conclusion:

[-x^2/3 cos(3x)] + [2x/9 sin(3x) + C]

However the answer I have been given is

[(2/27 - x^2/3) cos(3x)] + [2x/9 sin(3x) + C]

Similar but not quite right. Could anyone please tell me where I'm going wrong

Thanks

2. Hello, scott.b89!

Use integration by parts: .$\displaystyle I \;=\;\int x^2 \sin(3x)\,dx$

Could anyone please tell me where I'm going wrong?

It's hard to see your work from here,
but looks like you played the $\displaystyle Q\heartsuit$ instead of the $\displaystyle 7\spadesuit$
. . and lost a term . . .

By parts: .$\displaystyle \begin{array}{ccccccc}u &=& x^2 && dv &=&\sin(3x)\,dx \\ du &=& 2x\,dx && v &=&\text{-}\frac{1}{3}\cos(3x) \end{array}$

Then: .$\displaystyle I \;=\;\text{-}\tfrac{1}{3}x^2\cos(3x) + \tfrac{2}{3}\int x\cos(3x)\,dx$

By parts: .$\displaystyle \begin{array}{ccccccc} u &=& x && dv &=&\cos(3x)\,dx \\ du &=& dx && v &=&\frac{1}{3}\sin(3x) \end{array}$

Then: .$\displaystyle I \;=\;\text{-}\tfrac{1}{3}x^2\cos(3x) + \tfrac{2}{3}\bigg[\tfrac{1}{3}x\sin(3x) - \tfrac{1}{3}\int\sin(3x)\,dx \bigg]$

. . . . . $\displaystyle =\; \text{-}\tfrac{1}{3}x^2\cos(3x) + \tfrac{2}{9}x\sin(3x) - \tfrac{2}{9}\int\sin(3x)\,dx$

. . . . . $\displaystyle =\; \text{-}\tfrac{1}{3}x^2\cos(3x) + \tfrac{2}{9}x\sin(3x) + \tfrac{2}{27}\cos(3x) + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Then they factored $\displaystyle \cos(3x)$ from the first and third terms.

. . $\displaystyle \left(\tfrac{2}{27} - \tfrac{x^2}{3}\right)\cos(3x) + \tfrac{2}{9}x\sin(3x) + C$

I've seen only a few textbooks that do that.

Secondly, I see what you've done there!!! However,.. lol.

Sorry, but I'm not sure why that has been done, why is it that you need to take parts of the integral again? Could you give me a quick explanation please?

Thanks again!

4. Hi

You need to integrate twice by parts.
Each time you integrate by parts you lower the power of x by 1.
The first integration by part leads to $\displaystyle \int x\:\cos(3x)dx$ (not considering the constants).
The second one leads to $\displaystyle \int \sin(3x)dx$ which is easily integrable.

5. Aha,.. I see!

Cheers mate