JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Definite Integrals

Write the integrand as $\frac{f(x)}{x}$ with $f(x)=8\sin x-\sin 2x$. On $\left[\frac{\pi }{4},\frac{\pi }{3}\right]$ the derivative $f^{\prime }(x)=8\cos x-2\cos 2x$ stays positive, so $f$ increases from $f\left(\frac{\pi }{4}\right)=4\sqrt{2}-1\approx 4.66$ to $f\left(\frac{\pi }{3}\right)=\frac{7\sqrt{3}}{2}\approx 6.06$, while $x$ itself ranges only from $\frac{\pi }{4}\approx 0.785$ to $\frac{\pi }{3}\approx 1.047$. Dividing, the integrand lies between about $5.79$ and $5.93$ across the whole interval, so it is safely enclosed by $5<\frac{f(x)}{x}<8$. Multiplying these bounds by the interval width $\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$ gives $5\cdot \frac{\pi }{12}<I<8\cdot \frac{\pi }{12}$, that is $\frac{5\pi }{12}<I<\frac{2\pi }{3}$.