If $f(x)$ is a continuous function in $[0, \pi]$ such that $f(0)=f(\pi)=0$, then the value of $\int\limits_0^{\pi / 2}\left\{f(2 x)+f^{\prime \prime}(2 x)\right\} \sin x \cos x d x$ is equal to

0

Let $I=\int\limits_0^{\pi / 2}\left\{f(2 x)+f^{\prime \prime}(2 x)\right\} \sin x \cos x d x$. Then,

$I =\frac{1}{2} \int\limits_0^{\pi / 2}\left\{f(2 x)+f^{\prime \prime}(2 x)\right\} \sin 2 x d x$

$\Rightarrow I =\frac{1}{2} \int\limits_0^\pi\left\{f(t)+f^{\prime \prime}(t)\right\} \sin t d t$, where $t=2 x$

$\Rightarrow I =\frac{1}{2} \int\limits_0^\pi f(t) \sin t d t+\frac{1}{2} \int\limits_0^\pi f''(t) \sin t d t$

$\Rightarrow I=\frac{1}{2}\left[[-f(t) \cos t]_0^\pi+\int\limits_0^\pi f^{\prime}(t) \cos t d t+\left[-f^{\prime}(t) \sin t\right]_0^\pi - \int\limits_0^\pi f^{\prime}(t) \cos t d t\right]$

$\Rightarrow I=0$