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Question

Mathematics Question on integral

The value of ∫π6π31−sin2x dx∫_{\frac \pi6}^{\frac \pi3}\sqrt {1-sin2x\ dx} is

A

2−3+1\sqrt 2-\sqrt 3+1

B

22−3−12\sqrt 2-\sqrt 3-1

C

22+3−12\sqrt 2+\sqrt 3-1

D

2+3−1\sqrt 2+\sqrt 3-1

Answer

22−3−12\sqrt 2-\sqrt 3-1

Explanation

Solution

The correct option is (B): 22−3−12\sqrt 2-\sqrt 3-1.