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.
The reason why the lower and upper limits turn into the same value $0$ after changing variable is due to the lack of bijectivity of the substitution, even though the calculation is incorrect. Bijectivity in changing the variable is not a strong requirement, but it needs to be careful in calculation when a non-bijective substitution is applied.

Correct calculation using the same substitution is as follows: From $1-t^2 = x$, $t^2 = 1-x \ge 0$ and so
$$t = \begin{cases}
\sqrt{1-x} & \text{for $t \ge 0$,} \\
-\sqrt{1-x} & \text{for $t < 0$.}
\end{cases}$$
with
$$dt = \begin{cases}
-\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t > 0$} \\
\frac{1}{2\sqrt{1-x}}\,dx & \text{for $t < 0$.}
\end{cases}$$
Hence
\begin{align\*}
\int\_{-1}^{1}(1-t^2)\,dt 
 & = \int\_{-1}^{0} (1-t^2)\,dt + \int\_0^1 (1-t^2)\,dt \\
 & = \int\_0^1 x\left(\frac{1}{2\sqrt{1-x}}\right)\,dx 
 + \int\_1^0 x\left(\color{red}{-}\frac{1}{2\sqrt{1-x}}\right)\,dx\\
 & = \int\_0^1 \frac{x}{2\sqrt{1-x}}\,dx 
 + \int\_\color{red}{0}^\color{red}{1} \frac{x}{2\sqrt{1-x}}\,dx \\
 & = \int\_0^1 \frac{x}{\sqrt{1-x}}\,dx 
 = \lim\_{\varepsilon \rightarrow 1^-} \int\_0^{\varepsilon} \frac{x}{\sqrt{1-x}}\,dx\\
 & = \lim\_{\varepsilon \rightarrow 1^-} \left[\left.-\frac{2}{3}\sqrt{1-x}(x+2)\right|\_0^\varepsilon\right] \\
 & = \lim\_{\varepsilon \rightarrow 1^-} \left[\left(-\frac{2}{3}\sqrt{1-\varepsilon}(\varepsilon+2)\right) - \left(-\frac{4}{3}\right)\right] \\ 
 & = 0 - \left(-\frac{4}{3}\right) = \frac{4}{3}.
\end{align\*}