Integral of $$$- \sin{\left(3 x - 2 \right)}$$$

Find $$$\int \left(- \sin{\left(3 x - 2 \right)}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-1$$$ and $$$f{\left(x \right)} = \sin{\left(3 x - 2 \right)}$$$:

$${\color{red}{\int{\left(- \sin{\left(3 x - 2 \right)}\right)d x}}} = {\color{red}{\left(- \int{\sin{\left(3 x - 2 \right)} d x}\right)}}$$

Let $$$u=3 x - 2$$$.

Then $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

The integral can be rewritten as

$$- {\color{red}{\int{\sin{\left(3 x - 2 \right)} d x}}} = - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = - {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$

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

$$- \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{3} = - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{3}$$

Recall that $$$u=3 x - 2$$$:

$$\frac{\cos{\left({\color{red}{u}} \right)}}{3} = \frac{\cos{\left({\color{red}{\left(3 x - 2\right)}} \right)}}{3}$$

Therefore,

$$\int{\left(- \sin{\left(3 x - 2 \right)}\right)d x} = \frac{\cos{\left(3 x - 2 \right)}}{3}$$

Add the constant of integration:

$$\int{\left(- \sin{\left(3 x - 2 \right)}\right)d x} = \frac{\cos{\left(3 x - 2 \right)}}{3}+C$$

$$$\int \left(- \sin{\left(3 x - 2 \right)}\right)\, dx = \frac{\cos{\left(3 x - 2 \right)}}{3} + C$$$A