Question

∫cos(logx)÷x?

Answer 1

`I=sin(logx)+c`

Explanation:

Here,

`I=intcos(logx)/xdx=intcos(logx)*1/xdx`

We assume that , `logx=log_ex=lnx`

Subat. `logx=u=>1/xdx=du`

`I=intcosudu`

`=>I=sinu+c`

Subst. back ,`u=logx`

`I=sin(logx)+c`