Given, $I_1=\int_0^{\pi / 2} \frac{x}{\sin x} d x$
$$
I_2=\int_0^1 \frac{\tan ^{-1} x}{x} d x=\int_0^{\pi / 4} \frac{t}{\tan t} \cdot \sec ^2 t d t
$$
Put, $\tan ^{-1} x=t$
$$
\begin{aligned}
& x=\tan t \\
& d x=\sec ^2 t d t \\
& U . L \rightarrow \frac{\pi}{4}
\end{aligned}
$$
L. $L \rightarrow 0$
$$
\begin{aligned}
& =\int_0^{\pi / 4} t \cdot \frac{\cos t}{\sin t} \cdot \frac{1}{\cos ^2 t} d t=\int_0^{\pi / 4} \frac{t}{\sin t \cdot \cos t} d t \\
& =\int_0^{\pi / 4} \frac{2 t}{2 \sin t \cdot \cos t} d t=\int_0^{\pi / 4} \frac{2 t}{\sin 2 t} d t
\end{aligned}
$$
$$
\begin{aligned}
& \text { Now, put } 2 t=x \\
& 2 d t=d x \\
& d t=\frac{d x}{2} \\
& U . L \rightarrow \frac{\pi}{2} \\
&
\end{aligned}
$$

$$
\begin{aligned}
& L . L \rightarrow 0 \\
& =2 \int_0^{\pi / 2} \frac{x}{\sin x} d x \\
& I_2=2 \cdot I_1 \\
& \therefore \quad I_1: I_2=2: 1
\end{aligned}
$$
Hence, option (2) is correct.