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QJEEM23Ma95bc: JEE Main 2023 Mathematics - Class 12 JEEM23Ma95bc Solution: Let denote the greatest integer . Then 2 π ∫ π / 6 5 π / 6 ( 8…

Let [t][t] denote the greatest integer ≤t\leq t . Then 2π∫π/65π/6(8[cosecx]−5[cot⁡x])dx is equal to __________.

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amp; Let I=2π∫π65π6{8[cosec⁡x]−5[cot⁡x]}dxamp;=2π[8∫π65π6[cosec⁡x]dx−5∫π65π6[cot⁡x]dx]\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}

=2π[8∫π/65π/6dx−5{∫π/6π/4dx+∫π/4π/20.dx+∫π/23π/4(−1)dx++∫3π/45π/6(−2)dx}]]\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]

amp;=2π[8×(5π6−π6)−5{(π4−π6)−(3π4−π2)}−2(5π6−3π4)]amp;=2π[16π3+5π3]=14\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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