${\displaystyle \int \sin \left(\log x\right)dx=f\left(x\right)\left\{\sin \right(\log x)–\cos (\log x\left)\right\}+c}$ , then $f\left(x\right)=$

$I={\displaystyle \int \sin \left(\log x\right)dx}$

Put $\log x=t\Rightarrow x=e^t,dx=e^{t}dt$

$\therefore I={\displaystyle \int e^t\sin tdt}=\sin t{e}^t–{\displaystyle \int \left(\cos t{e}^t\right)dt}$

        $=\sin t{e}^t–\{\cos t{e}^t–{\displaystyle \int (–\sin t)e^t}dt\}$

       $=\sin t{e}^t–\cos t{e}^t–{\displaystyle \int \sin t{e}^t}dt$

$\Rightarrow 2I=(\sin t-\cos t)e^t+c$

$\Rightarrow I=\frac{1}{2}x$ $\left(\sin \left(\log x\right)–\cos \left(\log x\right)\right)+c$ .