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

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

Let $$$u=x y$$$.

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

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{y}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=x y$$$:

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

Therefore,

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

Add the constant of integration:

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

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