inte^(x)[tanx-log(cosx)]dx=...

`inte^(x)[tanx-log(cosx)]dx=`

A

`e^x log (sec x)+C`

B

`e^x log ("cosec")+c`

C

`e^x log (cos x)+C`

D

`e^x log ( sin x)+C `

Verified by Experts

The correct Answer is:

A

Similar Questions

Explore conceptually related problems

inte^x(tanx-logcosx)dx

inte^(x)(tanx+logsecx)dx=?

int(tanx)/(log(cosx))dx=

inte^(x)tanx(1+tanx)dx=

inte^(x)[secx+log(secx+tanx)]dx=?

int(tanx)/(1+cosx)dx=

If : int(f(x))/(log(cosx))dx=-log[log(cosx)]+c, then : f(x)=

int_(0)^(pi)log(1+cosx)dx=-pi(log2)

If int_(0)^((pi)/2)log(cosx)dx=-(pi)/2log2 , then int_(0)^((pi)/2)log(cosecx)dx=

int tanx dx=-log |cosx|+c