To find the value of the definite integral \(\rm\int_0^\pi x\ f(\sin x)\ dx\), we can use the properties of definite integrals.

Integral Property Application

Let the integral be denoted by I:

\(\rm I = \int_0^\pi x\ f(\sin x)\ dx\)

We use the property \(\rm \int_a^b f(x)\ dx = \int_a^b f(a+b-x)\ dx\). In this case, a = 0 and b = \pi.

Applying this property, we replace x with (\pi - x):

\(\rm I = \int_0^\pi (\pi - x)\ f(\sin(\pi - x))\ dx\)

We know that \(\rm \sin(\pi - x) = \sin x\). Substituting this back into the equation:

\(\rm I = \int_0^\pi (\pi - x)\ f(\sin x)\ dx\)

Now, we can split the integral:

\(\rm I = \int_0^\pi \pi\ f(\sin x)\ dx - \int_0^\pi x\ f(\sin x)\ dx\)

Notice that the second term is the original integral I. So, we have:

\(\rm I = \pi \int_0^\pi f(\sin x)\ dx - I\)

Adding I to both sides:

\(\rm 2I = \pi \int_0^\pi f(\sin x)\ dx\)

Dividing by 2:

\(\rm I = \frac{\pi}{2} \int_0^\pi f(\sin x)\ dx\)

Simplifying the Integral Limits

Now let's consider the integral \(\rm \int_0^\pi f(\sin x)\ dx\). We can use another property: \(\rm \int_0^{2a} f(x)\ dx = 2 \int_0^a f(x)\ dx\) if \(\rm f(2a-x) = f(x)\).

In our case, the integral is \(\rm \int_0^\pi f(\sin x)\ dx\). Here, 2a = \pi, so a = \pi/2.

Let's check the condition: \(\rm f(\sin(\pi - x))\) vs \(\rm f(\sin x)\).

Since \(\rm \sin(\pi - x) = \sin x\), we have \(\rm f(\sin(\pi - x)) = f(\sin x)\). The condition is satisfied.

Therefore, we can write:

\(\rm \int_0^\pi f(\sin x)\ dx = 2 \int_0^{\pi/2} f(\sin x)\ dx\)

Final Solution Derivation

Substitute this result back into our expression for I:

\(\rm I = \frac{\pi}{2} \left( 2 \int_0^{\pi/2} f(\sin x)\ dx \right)\)

The 2's cancel out:

\(\rm I = \pi \int_0^{\pi/2} f(\sin x)\ dx\)

This matches the first option.