Answer
$$\frac{1}{2}x(\sin (\ln x)+ \cos (\ln x))$$
Work Step by Step
Given $$\int \cos (\ln x)dx$$
Let $s=\ln x\ \Rightarrow ds=\dfrac{1}{x}dx\ \Rightarrow dx=e^sds$
$$I=\int \cos (\ln x)dx=\int e^s\cos sds$$
Let
\begin{align*}
u&=e^s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cos sds\\
u&=e^sds\ \ \ \ \ \ \ \ \ \ \ \ \ v=\sin s
\end{align*}
Then using integration by parts
\begin{align*}
I&=uv-\int vdu\\
&= e^s\sin s -\int e^s\sin sds \\
&=e^s\sin s-J
\end{align*}
where $J=\displaystyle\int e^s\sin sds$
Let
\begin{align*}
u&=e^s\ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\sin s ds\\
u&=e^sds\ \ \ \ \ \ \ \ \ \ \ \ \ v=-\cos s
\end{align*}
then
\begin{align*}
J&= -e^s\cos s +\int e^s\cos s ds\\
&=-e^s\cos s +I
\end{align*}
Hence
\begin{align*}
I &=e^s\sin s-J\\
&=e^s\sin s+e^s\cos s -I
\end{align*}
Hence
$$I=\frac{1}{2}e^s(\sin s+ \cos s)=\frac{1}{2}x(\sin (\ln x)+ \cos (\ln x))$$
