Dirac's delta, a simple integral

Solving $$\int_{-\infty}^\infty \delta(2x-\pi)\sin{x}dx$$ Should not be too hard. Dirac's delta assumes 1 at $x=\frac{\pi}{2}$ and zero everywhere else. So the answer is $\sin{\frac{\pi}{2}}=1$. But the correct answer is supposed to be $\frac{1}{2}$, why??

-
It's not true that "Dirac's delta assumes 1". If one insists (probably unreasonably) on assigning $\delta(x)$ a value at $x=0$, then that "value" has to be an $\infty$ so big that the integral $\int_{-\infty}^{+\infty}\delta(x)\,dx$ equals 1. –  Andreas Blass Jan 4 '13 at 20:52

Because you have $\delta(2 x - \pi)$ in the integral. You need to do a substitution with $u=2 x$ and this is where your factor of $1/2$ comes from.