Originally posted by Tom
What you did wrong was take the answer off a calculator.
Do a u-substitution:
then look up ∫sin2(u)du in an integral table. Don't forget to evaluate it at each limit.
You're not understading:
Let me give you all my work to alleaviate any confusion.
Show that A = (2/L)1/2
&psi(x) = A Sin(&pi x/L)
&psi2(x) = A2 Sin2(&pi x/L)
[inte]0L &psi2dx = 1
A2[inte]0L Sin2(&pi x/L) dx = 1
I forgot to resubsitute...
BTW: I only use a calculator/computer to check my work.