`int[sin(logx)+cos(logx)]dx`

Explanation

`"Let I=int [sin (log x)+cos (log x)]dx`
Put log x=t. Then, `x=e^(t) therefore dx=e^(t)dt`
`therefore I=int (sin t+cos t)e^(x)dt`
`"If" f(t)=sint, " then "f'(t)=cos t`
`therefore I=int e^(x)[f(t)+f'(t)]dt`
`therefore I=int e^(x)[f(t)+f'(t)]dt`
`=e^(t)f(t)+c=e^(t)sint+c`
`=x.sin (log x)+c`.