Consider
$ f(x)=8 \sin x-\sin 2 x$
$ f^{\prime}(x)=8 \sin x-2 \cos 2 x $
$ f^{\prime}(x)=-8 \sin x+4 \sin 2 x $
$ =-8 \sin x(1-\cos x)$
$ \therefore f^{\prime}(x)<0 x \in\left(\frac{\pi}{4}, \frac{\pi}{3}\right)$
$\therefore f ^{\prime}( x )$ is $\downarrow$ function
$f ^{\prime}\left(\frac{\pi}{3}\right)< f ^{\prime}( x )< f ^{\prime}\left(\frac{\pi}{4}\right)$
$ 5< f ^{\prime}( x )<\frac{8}{\sqrt{2}}$
$ 5< f ^{\prime}( x )<4 \sqrt{2}$
$ 5 x < f ( x )<4 \sqrt{2} x $
$5<\frac{ f ( x )}{ x }<4 \sqrt{2} $
$ \int\limits_{\pi / 4}^{\pi / 3} 5<\int \frac{ f ( x )}{ x }<\int\limits_{\pi / 4}^{\pi / 3} 4 \sqrt{2} $
$ \int\limits_{\pi / 4}^{\pi / 3} 5<\int \frac{8 \sin x -\sin 2 x}{ x }<\int\limits_{\pi / 4}^{\pi / 3} 4 \sqrt{2} $
$ \frac{5 \pi}{12}< I <\frac{\sqrt{2} \pi}{3}$