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

The given curve is $y=x+\sin x\cos x$

The given point is $x=\frac{\mathrm{\pi }}{2}$, $y=\frac{\mathrm{\pi }}{2}+0=\frac{\mathrm{\pi }}{2}$, it is $\left(\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

Differentiate the equation of the curve 

$\frac{dy}{dx}=1-{\sin }^{2}x+{\cos }^{2}x$

Let $m$ be the slope of the tangent to the curve at $\left(\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

$m=1-1+0=0$

Equation of normal is $x=k$, since the tangent is parallel to $x-$ axis

Therefore, the equation of normal is $x=\frac{\mathrm{\pi }}{2}\Rightarrow \boxed{\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{\pi }\mathbf{=}\mathbf{0}}$