This is the snippet of a problem from this PDF here. What I dont understand is why they retain the $Sin$ part for evaluation after integration when all that it is going to evaluate to is 0.
If I am not wrong, $Sin(n\pi) = 0$
All that should be left after integrating by parts is:
$$
[\frac{-2(x-1)Cos\frac{n\pi x}{2}}{n\pi}]_1^2
$$
Please correct me if I am wrong
The term you want to terminate is of the form $$f(x)|^2_1$$ (with $f(x)=(4/n^2\pi^2)\sin(n\pi x)/2$) which stands for $$f(2)-f(1)$$ It is true that $f(2)$ involves $\sin(n\pi)$ and is zero, but $f(1)$ involves $\sin(n\pi/2)$ and is not always zero.