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itn "cosec" x log ("cosec" x-cot x)dx=...

`itn "cosec" x log ("cosec" x-cot x)dx=`

A

`(1)/(2) log^(2)("cosec" x +cot )+c`

B

`(1)/(2) log^(2)("cosec" x -cot )+c`

C

`-(1)/(2) log^(2)("cosec" x -cot )+c`

D

`(1)/(3) log^(2)("cosec" x -cot )+c`

Text Solution

Verified by Experts

The correct Answer is:
B
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