Question: This problem is from BdMO. I am confused with the problem. Shouldn't it be infinity as there can be many people who have same birthday?

Answer: 
The substitution $$x=\arcsin t$$ is valid for $$-\pi/2\le x\le \pi/2$$ where $ \arcsin t$ is defined.
I will give you another example (much more *extreme*) so you can see plainly why is failing here.

If you consider $\int\_{-1}^1 (1-t^2) dt$ and you try to perform the substitution $1-t^2=x$, then $-2t dt = dx$, so $dt = \frac{-1}{2\sqrt{1-x}} dx$.

$$\int\_{-1}^1 (1-t^2) dt = \int\_{0}^0 x \frac{-dx}{2\sqrt{1-x}} = 0$$

Which makes no sense since the integral on the left is clearly non zero. The problem is that the substitution we made is not bijective on these intervals. The same happens with yours.