Step 1: Look at the integral.
We want $I=\int_0^{\pi/2}\frac{\sin x-\cos x}{1-\sin x\cos x}\,dx$. The tricky part is the numerator, which is a difference of sine and cosine.
Step 2: Notice a symmetry.
The limits run from $0$ to $\pi/2$. Whenever the limits are $0$ to $\pi/2$, swapping $x$ with $\pi/2-x$ turns sine into cosine and cosine into sine. So this swap is worth a try.
Step 3: Use the replacement property.
The rule is $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$. Here $a=\pi/2$, so replace every $x$ by $\pi/2-x$.
Step 4: Simplify the new form.
Since $\sin(\pi/2-x)=\cos x$ and $\cos(\pi/2-x)=\sin x$, we get $I=\int_0^{\pi/2}\frac{\cos x-\sin x}{1-\cos x\sin x}\,dx$. The bottom is unchanged, and the top has just flipped sign.
Step 5: Add the two forms.
The new integral is exactly $-I$, because the numerator is the negative of the original. So we have $I=-I$.
Step 6: Solve for I.
From $I=-I$ we get $2I=0$, hence $I=0$. So the answer is option (C).
\[ \boxed{\,I=0\,} \]