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Question

Mathematics Question on integral

I(x)=∫6dxsin2x(1+cot⁡x)2I(x)=\int\frac{6dx}{sin^2x(1+\cot x)^2}and I(0) then I (π2\frac{\pi}{2}) is equal to

A

21−933−3\frac{21-9\sqrt{3}}{3-\sqrt{3}}

B

21+933−3\frac{21+9\sqrt{3}}{3-\sqrt{3}}

C

213−3\frac{21}{3-\sqrt{3}}

D

3+33−3\frac{3+\sqrt{3}}{3-\sqrt{3}}

Answer

21−933−3\frac{21-9\sqrt{3}}{3-\sqrt{3}}

Explanation

Solution

The Correct answer is option is (A) : 21−933−3\frac{21-9\sqrt{3}}{3-\sqrt{3}}