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The value of int cos (logx)dx is...

The value of `int cos (logx)dx` is

A

`(1)/(2)[sin (logx)+cos (logx)]+C`

B

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

C

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

D

`(1)/(2)[sin(logx)-cos (logx)]+C`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `" "l=int cos (logx).1dx" …(i)"`
Use integral by parts,
`l=cos(logx).x -int[-sin (logx)].(1)/(x).xdx`
`=x.cos(logx)+int sin (logx).1dx`
Again on using integration by part,
`=x-cos(logx)+[sin(logx).x-int cos(logx).(1)/(x).xdx]+C`
`=x.cos(logx)+[x sin(logx)-int cos(logx)dx]+C`
`=x {sin(logx)+cos(logx)}-l+C" [from Eq. (i)]"`
`rArr" "l=(x)/(2){sin (logx)+cos(logx)}+C`
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