Step 1: Write down the integral.
We must evaluate \[ I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx \] A direct attack is messy, so we lean on a symmetry property.
Step 2: Recall the King's property.
For any nice function, \[ \int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx \] Here $a = \pi/2$, so we replace $x$ by $\frac{\pi}{2} - x$.
Step 3: Apply the swap.
Using $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$, the integral becomes \[ I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x}\,dx \] Call this the second form.
Step 4: Add the two forms.
Adding the original $I$ and this new $I$ lets the denominators line up: \[ 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx \]
Step 5: Simplify the integrand.
The fraction collapses to $1$, so \[ 2I = \int_0^{\pi/2} 1\,dx = \Big[x\Big]_0^{\pi/2} = \frac{\pi}{2} \]
Step 6: Solve for $I$.
Dividing by $2$ gives \[ I = \frac{\pi}{4} \] which is option (B).
\[ \boxed{I = \dfrac{\pi}{4}} \]