$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \mathrm{~d} x=\ldots$

Evaluate the integral: $I = \int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \, \mathrm{d} x$

Apply the transformation $x \mapsto \frac{\pi}{2} - x$, using the identity $\int_a^b f(x) \, \mathrm{d} x = \int_a^b f(a+b-x) \, \mathrm{d} x$:

$I = \int_0^{\frac{\pi}{2}} \frac{300 \cos x+100 \sin x}{\cos x+\sin x} \, \mathrm{d} x$

Add this to the original integral:

$2I = \int_0^{\frac{\pi}{2}} \frac{400 \sin x+400 \cos x}{\sin x+\cos x} \, \mathrm{d} x = 400 \int_0^{\frac{\pi}{2}} \, \mathrm{d} x$

Evaluate the integral:

$2I = 400 \cdot \frac{\pi}{2} = 200\pi$

Result: $I = 100\pi$

This corresponds to option A.