If \(\mathrm{I}=\int \sin (\log (x)) \mathrm{d} x\), then \(\mathrm{I}\) is given by

Let \(\log x=\mathrm{t} \Rightarrow x=\mathrm{e}^{\mathrm{t}}\) Differentiating w.r.t. t, we get \(\mathrm{d} x=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
\(\begin{aligned}
\therefore \quad \mathrm{I} & =\int \sin (\mathrm{t}) \mathrm{e}^{\mathrm{t}} \mathrm{dt} \\
& =\sin \mathrm{t} \mathrm{e}^{\mathrm{t}}-\int \cos \mathrm{t} \mathrm{e}^{\mathrm{t}} \mathrm{dt} \\
& =\sin \mathrm{t} \mathrm{e}^{\mathrm{t}}-\left[\cos \mathrm{t} \mathrm{e}^{\mathrm{t}}+\int \sin \mathrm{t} \mathrm{e}^{\mathrm{t}} \mathrm{dt}\right] \\
& =\sin \mathrm{t} \mathrm{e}^{\mathrm{t}}-\cos \mathrm{t} \mathrm{e}^{\mathrm{t}}-\mathrm{I}
\end{aligned}\)
\(\begin{array}{ll}\therefore & 2 \mathrm{I}=x \sin (\log x)-x \cos (\log x)+\mathrm{c} \\ \therefore \quad & \mathrm{I}=\frac{x}{2}(\sin (\log x)-\cos (\log x))+\mathrm{c}\end{array}\)