Let [t] denote the greatest integer ≤ t. Then 2 π ∫ _ π 6^ 5π 6(8[ cosecx]-5[ cotx])dx is equal to _______

Let,

$$I={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}(8[\mathrm{cosec}x]-5[\mathrm{cot}x])dx$$I=∫6π​65π​​(8[cosecx]−5[cotx])dx

$$\Rightarrow I=8{\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cosec}x]dx-5{\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cot}x]dx$$⇒I=8∫6π​65π​​[cosecx]dx−5∫6π​65π​​[cotx]dx

$\Rightarrow I=8{I}_{1}-5{I}_{2}$⇒I=8I1​−5I2​, where \displaystyle {I}_{1}=8{\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cosec}x]dx&{I}_{2}=5{\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cot}x]dx

Now solving,

$${I}_{1}={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cosec}x]dx$$I1​=∫6π​65π​​[cosecx]dx

$$\Rightarrow {I}_{1}={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}1dx=\frac{2\pi }{3}$$⇒I1​=∫6π​65π​​1dx=32π​

As when $x\in (\frac{\pi }{6},\frac{5\pi }{6}),\mathrm{cosec}x\in [1,2)$x∈(6π​,65π​),cosecx∈[1,2),

So, $[\mathrm{cosec}x]=1$[cosecx]=1

Now solving,

$${I}_{2}={\int }_{\frac{\pi }{6}}^{\frac{5\pi }{6}}[\mathrm{cot}x]dx$$I2​=∫6π​65π​​[cotx]dx

$$\Rightarrow {I}_{2}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}1dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}0dx+{\int }_{\frac{\pi }{2}}^{\frac{3\pi }{4}}(-1)dx+{\int }_{\frac{3\pi }{4}}^{\frac{5\pi }{6}}(-2)dx$$⇒I2​=∫6π​4π​​1dx+∫4π​2π​​0dx+∫2π​43π​​(−1)dx+∫43π​65π​​(−2)dx

$$\Rightarrow {I}_{2}=(\frac{\pi }{4}-\frac{\pi }{6})-(\frac{3\pi }{4}-\frac{\pi }{2})-2(\frac{5\pi }{6}-\frac{3\pi }{4})$$⇒I2​=(4π​−6π​)−(43π​−2π​)−2(65π​−43π​)

$$\Rightarrow {I}_{2}=-\frac{\pi }{3}$$⇒I2​=−3π​

Required value will be,

$$\frac{2}{\pi }I=\frac{2}{\pi }[8\times \frac{2\pi }{3}+5\times \frac{\pi }{3}]=14$$π2​I=π2​[8×32π​+5×3π​]=14