The equation of the normal to the curve $y=x+\sin x \cos x$ at $x=\frac{\pi}{2}$, is

$2 x=\pi$

When $x=\frac{\pi}{2}$, we have

$y=\frac{\pi}{2}+\sin \frac{\pi}{2} \cos \frac{\pi}{2}=\frac{\pi}{2}$

So, the coordinates of the point are $(\pi / 2, \pi / 2)$ Now,

$y=x+\sin x \cos x$

$\Rightarrow \quad \frac{d y}{d x}=1+\cos ^2 x-\sin ^2 x \Rightarrow\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}}=1+0-1=0$

This means that the tangent at $(\pi / 2, \pi / 2)$ is parallel to x-axis. So, the normal is parallel to y-axis and hence its equation is $x=\frac{\pi}{2}$.