1. Given equation

   We need to solve $f(x)=x+\int_0^{\pi/2} \sin x\,\cos y\,f(y)\,dy.$

2. Factor out terms independent of $y$

   Since $\sin x$ does not depend on $y$, we can write $f(x)=x+\sin x\int_0^{\pi/2} \cos y\,f(y)\,dy.$

   Let $I=\int_0^{\pi/2} \cos y\,f(y)\,dy.$

   Then $f(x)=x+I\sin x.$

3. Substitute this form back into the definition of $I$

   Using $f(y)=y+I\sin y$, $I=\int_0^{\pi/2} \cos y\,(y+I\sin y)\,dy.$

   So, $I=\int_0^{\pi/2} y\cos y\,dy+I\int_0^{\pi/2} \sin y\cos y\,dy.$

4. Evaluate the integrals

   First, $\int_0^{\pi/2} y\cos y\,dy.$

   Use integration by parts: $u=y,\quad dv=\cos y\,dy$ $du=dy,\quad v=\sin y.$

   Hence, $\int y\cos y\,dy=y\sin y-\int \sin y\,dy=y\sin y+\cos y.$

   Therefore,

   =\frac{\pi}{2}-1.$$ Next, $$\int_0^{\pi/2} \sin y\cos y\,dy=\frac12\int_0^{\pi/2} \sin 2y\,dy=\frac12.$$

5. Solve for $I$

   Thus, $I=\left(\frac{\pi}{2}-1\right)+\frac{I}{2}.$

   So, $\frac{I}{2}=\frac{\pi}{2}-1$ $I=\pi-2.$

6. Obtain $f(x)$

   Since $f(x)=x+I\sin x,$ we get $f(x)=x+(\pi-2)\sin x.$

7. Match with the options

   This is exactly Option D.

Comparison with stored answer: Stored correct answer is D, which matches our result.