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inte^x(tanx-logcosx)dx...

`inte^x(tanx-logcosx)dx`

A

`e^(x)tanx+C`

B

`e^(x)logcosx+C`

C

`e^(x)logsecx+C`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
`I=inte^(x){f(x)+f'(x)}dx," where "f(x)=logsecx`
`=e^(x)f(x)+C=e^(x)logsecx+C`.
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