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int ((1+tanx)/(1-tanx))dx...

`int ((1+tanx)/(1-tanx))dx`

A

`-log|cosx-sinx|+C`

B

`log|cosx-sinx|+C`

C

`log|cosx+sinx|+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`I=int((1+(sinx)/(cosx)))/((1-(sinx)/(cosx)))dx=int((cosx+sinx))/((cosx-sinx))dx`.
`=-int(dt)/(t),"where "(cosx-sinx)=t`
`=-log|t|+C=-log|cosx-sinx|+C`.
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