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

Find $$$\int \left(- 3 \sin{\left(3 t \right)}\right)\, dt$$$.

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

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

Let $$$u=3 t$$$.

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

The integral can be rewritten as

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

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=3 t$$$:

$$\cos{\left({\color{red}{u}} \right)} = \cos{\left({\color{red}{\left(3 t\right)}} \right)}$$

Therefore,

$$\int{\left(- 3 \sin{\left(3 t \right)}\right)d t} = \cos{\left(3 t \right)}$$

Add the constant of integration:

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

$$$\int \left(- 3 \sin{\left(3 t \right)}\right)\, dt = \cos{\left(3 t \right)} + C$$$A