Answer
$$\frac{1}{2} x(\sin (\ln x)- \cos (\ln x))+C$$
Work Step by Step
Given$$\int \sin (\ln x)dx $$
Let $$z=\ln x\ \ \ \ \ \ dz=\frac{1}{x}dx $$
Then $$ \int \sin (\ln x)dx =\int e^z\sin zdz $$
\begin{align*}
u &= \sin z \ \ \ \ \ \ \ \ \ dv= e^ zdz\\
du &= \cos z dz \ \ \ \ \ \ \ \ v= e^ z
\end{align*}
Then
\begin{align*}
\int \sin z e^{ z} d x &=(\sin z)\left(e^{ z}\right)-\int e^{ z} \cos z d z \\
&=e^{ z} \sin z- \int \cos z e^{ z} d z
\end{align*}
For the integral part again, let
\begin{align*}
u &= \cos z\ \ \ \ \ \ \ \ \ dv= e^ zdz\\
du &= -\sin zdz \ \ \ \ \ \ \ \ v= e^z
\end{align*}Then,
\begin{align*}
\int \sin z e^{z} d &=e^{z} \sin z-\left(e^{z} \cos z+ \int e^{z} \sin z d z\right)\\
\int \sin ze^{z} dz&=e^{z} \sin z- e^{z} \cos z- \int e^{z} \sin z dz\\
2 \int \sin z e^{z} d x&=e^{z} \sin z- e^{z} \cos z\\
\int \sin z e^{z} d z&=\frac{1}{2} e^{z}(\sin z- \cos z)+C
\end{align*}
Hence
$$\int \sin (\ln x)dx =\frac{1}{2} x(\sin (\ln x)- \cos (\ln x))+C $$
