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Question: The value of \(\int_{0}^{\pi/2\int(x - \pi/3)\cos e(x - \pi/.6)}\cos\) is...

The value of ∫0π/2∫(x−π/3)cos⁡e(x−π/.6)cos⁡\int_{0}^{\pi/2\int(x - \pi/3)\cos e(x - \pi/.6)}\cos is

A

2log3

B

-2log 3

C

log3

D

None of these

Answer

2log3

Explanation

Solution

∫0π/2∫(x−π/3)cos⁡e(x−π/6)cos⁡\int_{0}^{\pi/2\int(x - \pi/3)\cos e(x - \pi/6)}\cos

= 2∫0π/2sin⁡[(x−π6)−(x−π3)]sin⁡(x−π6).sin⁡(x−π3)dx2\int_{0}^{\pi/2}{\frac{\sin\left\lbrack \left( x - \frac{\pi}{6} \right) - \left( x - \frac{\pi}{3} \right) \right\rbrack}{\sin\left( x - \frac{\pi}{6} \right).\sin\left( x - \frac{\pi}{3} \right)}dx}

⇒2∫0π/2[cot⁡(x−π3)−cot⁡(x−π6)]dx2\int_{0}^{\pi/2}{\left\lbrack \cot\left( x - \frac{\pi}{3} \right) - \cot\left( x - \frac{\pi}{6} \right) \right\rbrack dx}

⇒ 2[log⁡sin⁡(x−π3)−log⁡sin⁡(x−π6)]0π/22\left\lbrack \log{\sin\left( x - \frac{\pi}{3} \right)} - \log{\sin\left( x - \frac{\pi}{6} \right)} \right\rbrack_{0}^{\pi/2}

⇒ 2[log⁡(sin⁡(x−π3)sin⁡(x−π6))]0π/2=2[log⁡(1232)−log⁡(3212)]2\left\lbrack \log\left( \frac{\sin\left( x - \frac{\pi}{3} \right)}{\sin\left( x - \frac{\pi}{6} \right)} \right) \right\rbrack_{0}^{\pi/2} = 2\left\lbrack \log\left( \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) - \log\left( \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \right) \right\rbrack⇒ 2[−log⁡3−log⁡3]2\left\lbrack - \log\sqrt{3} - \log\sqrt{3} \right\rbrack

⇒ −4log⁡3=−2log⁡3.- 4\log\sqrt{3} = - 2\log 3.