1. Given integral

$$I=\int_{\pi/4}^{\pi/3}\frac{8\sin x-\sin 2x}{x}\,dx$$

We need to estimate the value of $I$ and identify the correct interval.

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2. Simplify the numerator

Using $\sin 2x=2\sin x\cos x$,

$$8\sin x-\sin 2x=8\sin x-2\sin x\cos x=2\sin x(4-\cos x)$$

So,

$$I=\int_{\pi/4}^{\pi/3}\frac{2\sin x(4-\cos x)}{x}\,dx$$

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3. Estimate the integrand on the interval

For $x\in\left[\frac{\pi}{4},\frac{\pi}{3}\right]$:

• $\sin x$ is increasing, so $$\frac{1}{\sqrt2}\le \sin x\le \frac{\sqrt3}{2}$$
• $\cos x$ is decreasing, so $$\frac12\le \cos x\le \frac{1}{\sqrt2}$$ hence $$4-\frac{1}{\sqrt2}\le 4-\cos x\le \frac72$$
• Also, since $1/x$ is decreasing, $$\frac{3}{\pi}\le \frac1x\le \frac{4}{\pi}$$

Thus the integrand

$$f(x)=\frac{2\sin x(4-\cos x)}{x}$$

satisfies

$$f(x)\ge \frac{2\cdot \frac1{\sqrt2}\left(4-\frac1{\sqrt2}\right)}{x}$$

and

$$f(x)\le \frac{2\cdot \frac{\sqrt3}{2}\cdot \frac72}{x}=\frac{7\sqrt3}{2x}$$

But a sharper approach is to bound the factor $8\sin x-\sin 2x$ first.

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4. Bounds for $8\sin x-\sin 2x$

At $x=\pi/4$,

$$8\sin\frac\pi4-\sin\frac\pi2=8\cdot \frac1{\sqrt2}-1=4\sqrt2-1$$

At $x=\pi/3$,

$$8\sin\frac\pi3-\sin\frac{2\pi}3=8\cdot \frac{\sqrt3}{2}-\frac{\sqrt3}{2}=\frac{7\sqrt3}{2}$$

Now note that

$$\frac{d}{dx}(8\sin x-\sin 2x)=8\cos x-2\cos 2x$$

For $x\in\left[\frac\pi4,\frac\pi3\right]$, we have $\cos x>0$ and $\cos 2x\in[-1/2,0]$, so

$$8\cos x-2\cos 2x>0$$

Hence $8\sin x-\sin 2x$ is increasing on the interval. Therefore,

$$4\sqrt2-1\le 8\sin x-\sin 2x\le \frac{7\sqrt3}{2}$$

So,

$$(4\sqrt2-1)\int_{\pi/4}^{\pi/3}\frac{dx}{x} \le I\le \frac{7\sqrt3}{2}\int_{\pi/4}^{\pi/3}\frac{dx}{x}$$

That is,

$$(4\sqrt2-1)\ln\frac{\pi/3}{\pi/4} \le I\le \frac{7\sqrt3}{2}\ln\frac{\pi/3}{\pi/4}$$

Since

$$\ln\frac{\pi/3}{\pi/4}=\ln\frac43$$

we get

$$(4\sqrt2-1)\ln\frac43\le I\le \frac{7\sqrt3}{2}\ln\frac43$$

Numerically,

$$\ln\frac43\approx 0.28768$$

Lower bound:

$$(4\sqrt2-1)\ln\frac43\approx (5.6568-1)(0.28768)=4.6568\times 0.28768\approx 1.339$$

Upper bound:

$$\frac{7\sqrt3}{2}\ln\frac43\approx 6.062\times 0.28768\approx 1.744$$

So,

$$1.339<I<1.744$$

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5. Compare with the options

Let us compute the option endpoints numerically:

• $$$$

\quad \frac{3\pi}{4}\approx 2.356$$

• $$$$

\quad \frac{5\pi}{12}\approx 1.309$$

• $$$$

\quad \frac{\sqrt2}{3}\pi\approx 1.481$$

• $$$$

\quad \pi\approx 3.142$$

Our rough bounds give $1.339<I<1.744$, which already shows:

• Option B is false because $I>1.309$.
• Option D is false.

To distinguish between A and C, we estimate $I$ more accurately.

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6. Numerical estimation of the integral

Take

$$f(x)=\frac{8\sin x-\sin 2x}{x}$$

Evaluate at three points for Simpson's rule on $\left[\frac\pi4,\frac\pi3\right]$:

$$a=\frac\pi4,\quad b=\frac\pi3,\quad m=\frac{a+b}{2}=\frac{7\pi}{24}$$

Now,

$$f\left(\frac\pi4\right)=\frac{4\sqrt2-1}{\pi/4}=\frac{4(4\sqrt2-1)}{\pi}\approx 5.928$$

At $x=\frac{7\pi}{24}=52.5^\circ$,

$$\sin x\approx 0.79335,\quad \sin 2x=\sin 105^\circ\approx 0.96593$$

So,

$$f(m)\approx \frac{8(0.79335)-0.96593}{7\pi/24} =\frac{5.3809}{0.9163}\approx 5.872$$

Also,

f\left(\frac\pi3\right)=\frac{\frac{7\sqrt3}{2}}{\pi/3}=rac{21\sqrt3}{2\pi}\approx 5.790

Simpson's rule gives

$$I\approx \frac{b-a}{6}\left[f(a)+4f(m)+f(b)\right]$$

Now,

$$b-a=\frac\pi3-\frac\pi4=\frac\pi{12}$$

Hence,

$$I\approx \frac{\pi/12}{6}\left(5.928+4(5.872)+5.790\right)$$ $$=\frac{\pi}{72}(5.928+23.488+5.790)$$ $$=\frac{\pi}{72}(35.206)\approx 1.536$$

Thus,

$$I\approx 1.536$$

Now compare with the intervals:

• Option A: $\frac\pi2<I<\frac{3\pi}{4} \iff 1.571<I<2.356$ This is false since $I\approx 1.536<1.571$.

• Option C: $\frac{5\pi}{12}<I<\frac{\sqrt2}{3}\pi \iff 1.309<I<1.481$ This is also false since $I\approx 1.536>1.481$.

So neither A nor C fits the actual value.

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7. A clean analytic lower bound to confirm $I>\frac{\sqrt2}{3}\pi$

Since $f(x)=\dfrac{8\sin x-\sin 2x}{x}$ is decreasing only slightly and stays above its minimum value on the interval,

$$I>(b-a)\,f\left(\frac\pi3\right)$$

because $f(x)>f(\pi/3)$ for most of the interval, and numerically even the rectangle estimate gives

$$I>\frac\pi{12}\cdot 5.790\approx 1.516$$

Now,

$$\frac{\sqrt2}{3}\pi\approx 1.481$$

Hence,

$$I>1.516>1.481=\frac{\sqrt2}{3}\pi$$

Therefore option C cannot be correct.

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8. Conclusion

The actual value is approximately

$$I\approx 1.536$$

This does not lie in any of the listed intervals. In particular, the stored answer C is incorrect because

$$I>\frac{\sqrt2}{3}\pi.$$