Chapter 7 - Techniques of Integration - Review - Exercises - Page 578: 9

$$-\cos (\ln t)+c$$

Given $$\int \frac{\sin (\ln t)}{t} d t$$ Let $ u= \ln t\ \ \ \Rightarrow \ \ du=\frac{1}{t}dt $, then \begin{align*} \int \frac{\sin (\ln t)}{t} d t&=\int \sin (u) du\\ &=-\cos u+c\\ &= -\cos (\ln t)+c \end{align*}