# Thread: [SOLVED] Issue with a Integration by parts problem

1. ## [SOLVED] Issue with a Integration by parts problem

Hey,
I've run across a problem here that I can't seem to figure out. I need to Integrate a problem using the Integration by parts method.

$\displaystyle \int x^2\sin{(\pi x)} dx$

from there, I find:

$\displaystyle u=x^2$
$\displaystyle du=2x$
$\displaystyle dv=\sin{(\pi x)}dx$
$\displaystyle v=\frac{-1}{\pi}\cos{(\pi x)}$

and get:

$\displaystyle \frac{-1}{\pi}x^2 \cos{(\pi x)}-\int \frac{-2x}{\pi}\cos{(\pi x)}dx$

and then the answer I get is:

$\displaystyle \frac{-1}{\pi}x^2\cos{(\pi x)}+\frac{2x}{\pi^2}\sin{(\pi x)} + C$

But the book gives this as the answer:
$\displaystyle \frac{-1}{\pi}x^2\cos{(\pi x)}+\frac{2x}{\pi^2}\sin{(\pi x)}+\frac{2}{\pi^3}\cos{(\pi x)}+C$

I'm completely stumped as to where they got the last part (before the constant C) from.

Anyone?

2. you have to use parts twice ...

$\displaystyle u = x^2$
$\displaystyle du = 2x \, dx$

$\displaystyle dv = \sin(\pi x) \, dx$
$\displaystyle v = -\frac{1}{\pi} \cos(\pi x)$

$\displaystyle \int x^2 \sin(\pi x) \, dx = -\frac{x^2}{\pi} \cos(\pi x) + \frac{2}{\pi} \int x \cos(\pi x) \, dx$

using parts a second time for the last integral expression ...

$\displaystyle u = x$
$\displaystyle du = dx$

$\displaystyle dv = \cos(\pi x) \, dx$
$\displaystyle v = \frac{1}{\pi} \sin(\pi x)$

$\displaystyle \int x^2 \sin(\pi x) \, dx = -\frac{x^2}{\pi} \cos(\pi x) + \frac{2x}{\pi^2} \sin(\pi x) - \frac{2}{\pi^2} \int \sin( \pi x) \, dx =$

$\displaystyle -\frac{x^2}{\pi} \cos(\pi x) + \frac{2x}{\pi^2} \sin(\pi x) + \frac{2}{\pi^3} \cos(\pi x) + C$

3. Ah, I didn't know you could build it like that.

Thanks!

4. Originally Posted by Coco87
Sorry to bring this thread back up, but I figured this would be better than starting a new one. I have run into another brick wall that is driving me insane (see following picture: )

It seems simple, yet I'm not getting the answer in the book, but rather I'm looping through the Integration By Parts method over and over again.

$\displaystyle \int{\ln{(2x+1)}dx}$

I set:

$\displaystyle u=\ln{(2x+1)}$ $\displaystyle dv=dx$
$\displaystyle du=\frac{2}{2x+1}$ $\displaystyle v=x$

and get:

$\displaystyle x\ln{(2x+1)}-\int{\frac{2x}{2x+1}}$

and from there I integrate by partial fractions:

$\displaystyle 2\int{x\frac{1}{2x+1}}$

$\displaystyle u=\frac{1}{2x+1}$ $\displaystyle dv=x$
$\displaystyle du=\frac{2}{(2x+1)^2}$ $\displaystyle v=\frac{x^2}{2}$

then I get:

$\displaystyle 2[\frac{x^2}{2(2x+1)}-\int{\frac{x^2}{(2x+1)^2}}]$

I might be wrong, but it seems like and endless loop to me

Could anyone help me out?
$\displaystyle \int\frac{2x}{2x+1}dx=\int\underset{1}{\underbrace {\frac{2x+1}{2x+1}}}dx-\int\frac{1}{2x+1}dx$

.........._$\displaystyle =x-\frac{1}{2}\ln|2x+1|$

5. Originally Posted by bkarpuz

.........._$\displaystyle =x-\frac{1}{2}\ln|2x+1|$
$\displaystyle +k.$

6. Originally Posted by Krizalid
$\displaystyle +k.$
sure!