5.5 The Substitution Rule/17

$\int {\frac {\sin {(\ln {(x))}}}{x}}dx$

${\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}$

${\begin{aligned}\int {\frac {\sin {(\ln {(x))}}}{x}}dx&=\int {\frac {1}{x}}\sin(\ln {(x)})dx=\int \left({\frac {1}{x}}dx\right)\sin {(\ln {(x)})}\\[2ex]&=\int (du)\sin {(u)}=\int \sin {(u)}du\\[2ex]&=-\cos {(u)}+C\\[2ex]&=-\cos {(\ln {(x)})}+C\end{aligned}}$