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int cos (loge x)dx is equal to...

`int cos (log_e x)dx` is equal to

A

`x[cos (log_e x)+sin(log_e x)]+C`

B

`(x)/(2)[cos(log_e x)+sin(log_e x)]+C`

C

`x[cos(log_e x)-sin(log_e x)]+C`

D

`(x)/(2) [sin (log_e x)-cos (log_e x)]+C`.

Text Solution

Verified by Experts

The correct Answer is:
B
`I=x cos (log_e x)+int sin (log_e)(1)/(x).x.dx+C`.
`I= x cos (log_e x)+[ x sin log_e x -int cos (log_e x)((1)/(x)).x dx]`
`I=x cos (log_e x)+x sin (log_(e) x)-I+C`.
`I=(x)/(2) (cos (log_e x)+sin(log_e x)+C`.
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