"Let $I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x$. Then"

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I comes out around 1.536 which is not satisfied by any given options.

$$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $$

$${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $$

${{\sin x} \over x}$ is decreasing in $\left( {{\pi \over 4},{\pi \over 3}} \right)$ so it attains maximum at $x = {\pi \over 4}$

$$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $$

$I > \sqrt 3 - {\pi \over 6}$

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Let $I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x$. Then