∫sin(logx)dx=f(x){sin(logx)–cos(logx)}+c , then f(x)=

I=∫sin(logx)dx Put logx=t  ⇒  x=et,  dx=etdt ∴   I=∫etsintdt=sintet–∫(costet)dt         =sintet–{costet–∫(–sint)etdt}        =sintet–costet–∫sintetdt ⇒ 2I=(sint−cost)et+c ⇒ I=12x (sin(logx)–cos(logx))+c .