[SOLVED] Issue with a Integration by parts problem

**Coco87** · September 22nd 2008, 06:02 PM

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.

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

from there, I find:

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

and get:

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

and then the answer I get is:

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

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

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

Anyone?

**skeeter** · September 22nd 2008, 06:20 PM

you have to use parts twice ...

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

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

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

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

$u = x$
$du = dx$

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

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

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

**Coco87** · September 22nd 2008, 06:42 PM

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

Thanks!

**bkarpuz** · September 23rd 2008, 08:01 AM

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.

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

I set:

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

and get:

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

and from there I integrate by partial fractions:

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

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

then I get:

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

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

Could anyone help me out?

$\int\frac{2x}{2x+1}dx=\int\underset{1}{\underbrace {\frac{2x+1}{2x+1}}}dx-\int\frac{1}{2x+1}dx$

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

**Krizalid** · September 23rd 2008, 08:43 AM

Originally Posted by bkarpuz

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

$+k.$

**bkarpuz** · September 23rd 2008, 08:53 AM

Originally Posted by Krizalid

$+k.$

sure!

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