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int(secx."cosec"x)/(2cotx-secx"cosec x")...

`int(secx."cosec"x)/(2cotx-secx"cosec x")dx` is equal to

A

`(1)/(2)ln|sec2x+tan2x|+C`

B

`ln|secx+"cosec x"|+C`

C

`ln |secx+Tanx|+C`

D

`(1)/(2)ln|secx+" cosec x"|+C`

Text Solution

Verified by Experts

The correct Answer is:
A
`I=int(sec x "cosec x")/(2cot x -sec x" cosec "x)dx`
`=int(dx)/(2cos^(2)x-1)`
`=intsec2xdx`
`=(1)/(2)ln|sec 2x+tan2x_+C`
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