The function $f\left(x\right),$ that satisfies the condition $f\left(x\right)=x+{\int }_0^{\pi /2}\sin x\cos yf\left(y\right)\mathrm{d}y,$ is :

Given the function $f\left(x\right)=x+{\int }_0^{\frac{\pi }{2}}\sin x·\cos yf\left(y\right)dy$

$f\left(x\right)=x+k\sin x ...\left(i\right)$

$f\left(y\right)=y+k\sin y ...\left(ii\right)$

$k={\int }_0^{\frac{\pi }{2}}\cos yf\left(y\right)dy$

From equations $\left(i\right) \& \left(ii\right)$

$f\left(x\right)=x+{\int }_0^{\frac{\pi }{2}}\sin x\cos y\left(y+k\sin y\right)dy$

$f\left(x\right)=x+\sin x{\int }_0^{\frac{\pi }{2}}y\cos ydy+\frac{k}{2}\sin x{\int }_0^{\frac{\pi }{2}}\sin 2ydy$

$f\left(x\right)=x+\sin x\left[y{\int }_0^{\frac{\pi }{2}}\cos ydy-{\int }_0^{\frac{\pi }{2}}\left\{\frac{d}{dy}y{\int }_0^{\frac{\pi }{2}}\cos ydy\right\}dy\right]+\frac{k}{2}\sin x{\int }_0^{\frac{\pi }{2}}\sin 2ydy$

$f\left(x\right)=x+\sin x\left[\frac{\pi }{2}-1\right]+\frac{k}{2}\sin x\times \left(1\right)$

$f\left(x\right)=x+\frac{(\pi -2)\sin x}{2}+\frac{k\sin x}{2} ...\left(iii\right)$

from $\left(i\right)$ and $\left(iii\right)$

$\mathrm{k}=\frac{\pi }{2}-1+\frac{\mathrm{k}}{2}$

$\mathrm{k}=\pi -2$

Hence,  $f\left(x\right)=x+(\pi -2)\sin x$