If $f \left(x\right)$ is a continuous function in $[0, \mathrm{\pi }]$ such that  $f\left(0\right)=f\left(\pi \right)=0$, then the value of  ${\int }_0^{\pi /2} \left\{f\left(2x\right)+f^{\prime \prime }\left(2x\right)\right\}\sin x\cos xdx$ is equal to

Let $I={\int }_0^{\pi /2} \left\{f\left(2x\right)+f^{\prime \prime }\left(2x\right)\right\}\sin x\cos xdx.$   Therefore

$I=\frac{1}{2}{\int }_0^{\pi /2} \left\{f\left(2x\right)+f^{\prime \prime }\left(2x\right)\right\}\sin 2xdx$

$\Rightarrow I=\frac{1}{2}{\int }_0^{\pi } \left\{f\left(t\right)+f^{\prime \prime }\left(t\right)\right\}\sin tdt$, where $t=2x$

$\begin{array}{l}\Rightarrow I=\frac{1}{2}{\int }_0^{\pi } f\left(t\right)\sin tdt+\frac{1}{2}{\int }_0^{\pi } f^{\prime \prime }\left(t\right)\sin tdt\\ \Rightarrow I=\frac{1}{2}\left[[-f(t)\cos t{]}_0^{\pi }+{\int }_0^{\pi } f^{\mathrm{\prime }}(t)\cos tdt+{\left[-f^{\mathrm{\prime }}\left(t\right)\sin t\right]}_0^{\pi }\right.\\ \\ \Rightarrow I=0\\ \end{array}\left.-{\int }_0^{\pi } f^{\mathrm{\prime }}\left(t\right)\cos tdt\right]$