Expected values in infinite square well

[Aikon]

Ok...this must sound stupid, because i didn't found answer on the web and on my books...but i am having trouble with the infinite square well.
I want to calculate <x>.
V(x)=0 for 0<=x<=a
$<x>=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx$
Doing integration by parts i got to:
$\frac{2}{a}\left[\frac{a^2}{2n}-\int^a_0\frac{a}{2n}dx\right]=0$
What i am doing wrong?
Thank you

 

[mfb]

I would expect an error in your integration, as the first expression is not zero and the following one is.

 

[tadchem]

$<x>=\frac{2}{a}\int^{a}_{0} x \sin^2(\frac{n\pi}{a}x)dx$

set
$u=\frac{n\pi}{a}x$

look up
$\int u \sin^2(u)du$
in a Table of Integrals to find
$\frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8}$

 

[Aikon]

set
$u=\frac{n\pi}{a}x$

look up
$\int u \sin^2(u)du$
in a Table of Integrals to find
$\frac{u^2}{4}-\frac{u\sin(2u)}{4}-\frac{cos(2u)}{8}$

Yeah, i got it...it is easier to use table of integrals.
it gives the expected $<x>=\frac{a}{2}$ for any value of n.
thanks