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

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

Let $$$u=\frac{x}{k}$$$.

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=k$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$${\color{red}{\int{k \sin{\left(u \right)} d u}}} = {\color{red}{k \int{\sin{\left(u \right)} d u}}}$$

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

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

Recall that $$$u=\frac{x}{k}$$$:

$$- k \cos{\left({\color{red}{u}} \right)} = - k \cos{\left({\color{red}{\frac{x}{k}}} \right)}$$

Therefore,

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

Add the constant of integration:

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

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