I = ∫ _ (π)/(4) ^ (π)/(3) ((8 x - 2x)/(x))dx . Then

Let, $f\left(x\right) = 8\sin x - \sin 2x$

$⇒ {f}^{'}\left(x\right) = 8\cos x - 2\cos 2x$

⇒ {f}^{' '}\left(x\right) = - 8\sin x + 4\sin 2x

⇒ {f}^{' '}\left(x\right) = - 8\sin x\left(1 - \cos x\right)

So, ${f}^{'}\left(x\right) < 0\text{when}x \in \left(\dfrac{\pi}{4}, \dfrac{\pi}{3}\right)$

Hence ${f}^{'}\left(x\right)$ is decreasing function in given domain,

Now, ${f}^{'}\left(\dfrac{\pi}{3}\right) < {f}^{'}\left(x\right) < {f}^{'}\left(\dfrac{\pi}{4}\right)$

$⇒ 5 < {f}^{'}\left(x\right) < \dfrac{8}{\sqrt{2}}$

$⇒ 5 < {f}^{'}\left(x\right) < 4\sqrt{2}$

Now integrating all with respect to $dx$

$∫ 5dx < ∫ {f}^{'}\left(x\right)dx < ∫ 4\sqrt{2}dx$

$⇒ 5x < f\left(x\right) < 4\sqrt{2}x$

$⇒ 5 < \dfrac{f\left(x\right)}{x} < 4\sqrt{2}$

Again integrating with ${∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}}$

We get, ${∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{4}}5 < ∫ \dfrac{f\left(x\right)}{x} < {∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}}4\sqrt{2}$

$⇒ {∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}}5 < ∫ \dfrac{8\sin x - \sin 2x}{x} < {∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}}4\sqrt{2}$

$⇒ \dfrac{5\pi}{12} < I < \dfrac{\sqrt{2}\pi}{3}$