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

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

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

Then $$$du=\left(x + y\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

Therefore,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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