Let I=∫π/4π/3 8sin⁡x−sin⁡2xxdx. Then

Considerf(x)=8sinx−sin2xf'(x)=8cosx−2cos2xf''(x)=−8sinx+4sin2x=−8sinx(1−cosx)∴f′′(x)<0,x∈π4,π3∴f ′(x) is ↓ function f′π3<f′(x)<f′π45<f ′(x)<825<f ′(x)<425x<f(x)<42x5<f(x)x<42∫π/4π/3 5<∫π/4π/3f(x)x<∫π/4π/3 42∫π/4π/3 5<∫π/4π/38sinx−sin2xx<∫π/4π/3 425π12<I<2π3

Let I=∫π/4π/3 8sin⁡x−sin⁡2xxdx. Then

Considerf(x)=8sinx−sin2xf'(x)=8cosx−2cos2xf''(x)=−8sinx+4sin2x=−8sinx(1−cosx)∴f′′(x)<0,x∈π4,π3∴f ′(x) is ↓ function f′π3<f′(x)<f′π45<f ′(x)<825<f ′(x)<425x<f(x)<42x5<f(x)x<42∫π/4π/3 5<∫π/4π/3f(x)x<∫π/4π/3 42∫π/4π/3 5<∫π/4π/38sinx−sin2xx<∫π/4π/3 425π12<I<2π3

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Let $I = \int_{π / 4}^{π / 3} (\frac{8 \sin x - \sin 2 x}{x}) dx$ . Then

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$\frac{π}{2} < I < \frac{3 π}{4}$

$\frac{π}{5} < I < \frac{5 π}{12}$

$\frac{5 π}{12} < I < \frac{\sqrt{2}}{3} π$

$\frac{3 π}{4} < I < π$

answer is C.

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Detailed Solution

Consider

$\begin{array}{l} f (x) = 8 sinx - \sin 2 x \\ f^{'} (x) = 8 cosx - 2 \cos 2 x \\ f^{''} (x) = - 8 sinx + 4 \sin 2 x \\ = - 8 sinx (1 - cosx) \\ ∴ f^{''} (x) < 0, x \in (\frac{π}{4}, \frac{π}{3}) \\ ∴ f^{'} (x) is ↓ function \\ f^{'} (\frac{π}{3}) < f^{'} (x) < f^{'} (\frac{π}{4}) \\ 5 < f^{'} (x) < \frac{8}{\sqrt{2}} \end{array}$

$\begin{array}{l} 5 < f^{'} (x) < 4 \sqrt{2} \\ 5 x < f (x) < 4 \sqrt{2} x \\ 5 < \frac{f (x)}{x} < 4 \sqrt{2} \\ \int_{π / 4}^{π / 3} 5 < \int_{π / 4}^{π / 3} \frac{f (x)}{x} < \int_{π / 4}^{π / 3} 4 \sqrt{2} \\ \int_{π / 4}^{π / 3} 5 < \int_{π / 4}^{π / 3} \frac{8 sinx - \sin 2 x}{x} < \int_{π / 4}^{π / 3} 4 \sqrt{2} \\ \frac{5 π}{12} < I < \frac{\sqrt{2} π}{3} \end{array}$