Answer
$$\int\sin(\ln x)dx=\frac{x}{2}(\sin\ln x-\cos\ln x)+C$$
Work Step by Step
$$A=\int\sin(\ln x)dx$$
Let $a=\ln x$, which means $x=e^a$
Then $da=\frac{1}{x}dx=\frac{1}{e^a}dx$. So $dx=e^ada$
$$A=\int\sin a\times e^ada=\int e^a\sin ada$$
Let $u=\sin a$ and $dv=e^ada$
We will have $du=\cos ada$ and $v=e^a$
Apply the formula $\int udv=uv-\int vdu$: $$A=e^a\sin a-\int e^a\cos ada$$
We carry out integration by parts one more time.
Let $u=\cos a$ and $dv=e^ada$
We will have $du=-\sin ada$ and $v=e^a$
Apply the formula $\int udv=uv-\int vdu$: $$A=e^a\sin a-\Big(e^a\cos a-\int -e^a\sin ada\Big)$$ $$A=e^a\sin a-\Big(e^a\cos a+\int e^a\sin ada\Big)$$
We notice that $\int e^a\sin ada$ is exactly the given integral to solve, which means it equals $A$.
$$A=e^a\sin a-\Big(e^a\cos a+A\Big)$$ $$A=e^a\sin a-e^a\cos a-A$$ $$2A=e^a\sin a-e^a\cos a+C=e^a(\sin a-\cos a)+C$$ $$A=\frac{e^a}{2}(\sin a-\cos a)+C$$
Replace $a$ back to $\ln x$: $$A=\frac{e^{\ln x}}{2}(\sin\ln x-\cos\ln x)+C$$ $$A=\frac{x}{2}(\sin\ln x-\cos\ln x)+C$$
