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inttanxdx=?...

`inttanxdx=?`

A

`log|cosx|+C`

B

`-log|cosx|+C`

C

`log|sinx|+C`

D

`-log|sinx|+C`

Text Solution

Verified by Experts

The correct Answer is:
B

`I=int(sinx)/(cosx)dx=int(-dt)/(t)`, where cos x=t.
`=-log|t|+C=-log|cosx|+C`.
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