The equation of the normal to the curve

Given that the curve $y=x+\sin x \cos x$

Differentiating both the sides w.r.t. $x$,

$\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$

Now,

Slope of the tangent $\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}=\frac{\pi}{2}\right)=1+\cos ^{2} \frac{\pi}{2}-\sin ^{2} \frac{\pi}{2}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1-1+0=0$

When $\mathrm{x}=\frac{\pi}{2}, \mathrm{y}=\frac{\pi}{2}$

Equation of the normal:

$\left(y-y_{1}\right)=\frac{-1}{\text { Slope of tangent }}\left(x-x_{1}\right)$

$\Rightarrow\left(y-\frac{\pi}{2}\right)=\frac{-1}{0}\left(x-\frac{\pi}{2}\right)$

$\Rightarrow 2 x=\pi$

Hence option D is correct.