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

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

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

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

The integral becomes

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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