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u = ln ( x ) d u = 1 x d x {\displaystyle {\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}}
∫ sin ( ln ( x ) ) x d x = ∫ 1 x sin ( ln ( x ) ) d x = ∫ ( 1 x d x ) sin ( ln ( x ) ) = ∫ ( d u ) sin ( u ) = ∫ sin ( u ) d u = − cos ( u ) + C = − cos ( ln ( x ) ) + C {\displaystyle {\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}}}
![{\displaystyle {\begin{aligned}u&=\ln(x)\\[2ex]du&={\frac {1}{x}}dx\\[2ex]\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/2dcd1cacb47fe0ded9f0cb8411297ff136d61465)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/92ec083cbb58cbf42efe622bfc22fd4c76bb337f)