Answer
$-\cos(\ln t)+C$
Work Step by Step
$\displaystyle \int\frac{\sin(\ln t)}{t}dt=\left[\begin{array}{l}
u=\ln t\\
du=dt/t
\end{array}\right]$
$=\displaystyle \int\sin udu$
$=-\cos u+C$
$=-\cos(\ln t)+C$
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