JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 18)
Let $$[t]$$ denote the greatest integer $$\leq t$$. Then $$\frac{2}{\pi} \int_\limits{\pi / 6}^{5 \pi / 6}(8[\operatorname{cosec} x]-5[\cot x]) d x$$ is equal to __________.
Answer
14
Explanation
$$
\begin{aligned}
& \text { Let } \mathrm{I}=\frac{2}{\pi} \int\limits_{\frac{\pi}{6}}^{ \frac{5\pi}{6}}\{8[\operatorname{cosec} x]-5[\cot x]\} d x \\\\
& =\frac{2}{\pi}\left[8 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\operatorname{cosec} x] d x-5 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\cot x] d x\right]
\end{aligned}
$$
$$ \left.\begin{array}{r} =\frac{2}{\pi}\left[8 \int\limits_{\pi / 6}^{5 \pi / 6} d x-5\left\{\int\limits_{\pi / 6}^{\pi / 4} d x+\int\limits_{\pi / 4}^{\pi / 2} 0 . d x+\int\limits_{\pi / 2}^{3 \pi / 4}(-1) d x+\right.\right. \left.\left.+\int\limits_{3 \pi / 4}^{5 \pi / 6}(-2) d x\right\}\right] \end{array}\right] $$
$$ \begin{aligned} & =\frac{2}{\pi}\left[8 \times\left(\frac{5 \pi}{6} - \frac{\pi}{6}\right)-5\left\{\left(\frac{\pi}{4}-\frac{\pi}{6}\right)-\left(\frac{3 \pi}{4}-\frac{\pi}{2}\right)\right\}\right.\left.-2\left(\frac{5 \pi}{6}-\frac{3 \pi}{4}\right)\right] \\\\ & =\frac{2}{\pi}\left[\frac{16 \pi}{3}+\frac{5 \pi}{3}\right]=14 \end{aligned} $$
$$ \left.\begin{array}{r} =\frac{2}{\pi}\left[8 \int\limits_{\pi / 6}^{5 \pi / 6} d x-5\left\{\int\limits_{\pi / 6}^{\pi / 4} d x+\int\limits_{\pi / 4}^{\pi / 2} 0 . d x+\int\limits_{\pi / 2}^{3 \pi / 4}(-1) d x+\right.\right. \left.\left.+\int\limits_{3 \pi / 4}^{5 \pi / 6}(-2) d x\right\}\right] \end{array}\right] $$
$$ \begin{aligned} & =\frac{2}{\pi}\left[8 \times\left(\frac{5 \pi}{6} - \frac{\pi}{6}\right)-5\left\{\left(\frac{\pi}{4}-\frac{\pi}{6}\right)-\left(\frac{3 \pi}{4}-\frac{\pi}{2}\right)\right\}\right.\left.-2\left(\frac{5 \pi}{6}-\frac{3 \pi}{4}\right)\right] \\\\ & =\frac{2}{\pi}\left[\frac{16 \pi}{3}+\frac{5 \pi}{3}\right]=14 \end{aligned} $$
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