Integral of $$$\frac{r \sin{\left(\ln\left(x\right) \right)}}{x}$$$ with respect to $$$x$$$

Find $$$\int \frac{r \sin{\left(\ln\left(x\right) \right)}}{x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=r$$$ and $$$f{\left(x \right)} = \frac{\sin{\left(\ln{\left(x \right)} \right)}}{x}$$$:

$${\color{red}{\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}} = {\color{red}{r \int{\frac{\sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}}$$

Let $$$u=\ln{\left(x \right)}$$$.

Then $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (steps can be seen »), and we have that $$$\frac{dx}{x} = du$$$.

So,

$$r {\color{red}{\int{\frac{\sin{\left(\ln{\left(x \right)} \right)}}{x} d x}}} = r {\color{red}{\int{\sin{\left(u \right)} d u}}}$$

The integral of the sine is $$$\int{\sin{\left(u \right)} d u} = - \cos{\left(u \right)}$$$:

$$r {\color{red}{\int{\sin{\left(u \right)} d u}}} = r {\color{red}{\left(- \cos{\left(u \right)}\right)}}$$

Recall that $$$u=\ln{\left(x \right)}$$$:

$$- r \cos{\left({\color{red}{u}} \right)} = - r \cos{\left({\color{red}{\ln{\left(x \right)}}} \right)}$$

Therefore,

$$\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x} = - r \cos{\left(\ln{\left(x \right)} \right)}$$

Add the constant of integration:

$$\int{\frac{r \sin{\left(\ln{\left(x \right)} \right)}}{x} d x} = - r \cos{\left(\ln{\left(x \right)} \right)}+C$$

$$$\int \frac{r \sin{\left(\ln\left(x\right) \right)}}{x}\, dx = - r \cos{\left(\ln\left(x\right) \right)} + C$$$A