The function f(x),

$f(x)=x+\int_{0}^{\pi / 2} \sin x \cos y f(y) d y$

$f(x)=x+\sin x \underbrace{\int_{0}^{\pi / 2} \cos y f(y) d y}_{K}$

$\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{x}+\mathrm{K} \sin \mathrm{x}$

$\Rightarrow \mathrm{f}(\mathrm{y})=\mathrm{y}+\mathrm{K} \sin \mathrm{y}$

Now $\mathrm{K}=\int_{0}^{\pi / 2} \cos \mathrm{y}(\mathrm{y}+\mathrm{K} \sin \mathrm{y}) \mathrm{dy}$

$\mathrm{K}=\int_{0}^{\pi / 2} \mathrm{y} \cos \mathrm{dy}+\int_{\text {Apply IBP }}^{\pi / 2} \cos \mathrm{y} \sin \mathrm{ydy}_{\text {Put siny } \mathrm{y}=\mathrm{t}}$

$\mathrm{K}=(\mathrm{y} \sin \mathrm{y})_{0}^{\pi / 2}-\int_{0}^{\pi / 2} \sin \mathrm{dy}+\mathrm{K} \int_{0}^{1} \mathrm{t} \mathrm{dt}$

$\Rightarrow \mathrm{K}=\frac{\pi}{2}-1+\mathrm{K}\left(\frac{1}{2}\right)$

$\Rightarrow \mathrm{K}=\pi-2$

So $f(x)=x+(\pi-2) \sin x$

Option (4)