The value of `int cos (logx)dx` isA. `(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`

The value of `int cos (logx)dx` isA. `(1)/(2)[sin (logx)+cos (logx)]+C

1.	The value of `int cos (logx)dx` isA. `(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`
Answer» Correct Answer - 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`