Integral of $$$6 \cos{\left(3 t \right)}$$$

Find $$$\int 6 \cos{\left(3 t \right)}\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=6$$$ and $$$f{\left(t \right)} = \cos{\left(3 t \right)}$$$:

$${\color{red}{\int{6 \cos{\left(3 t \right)} d t}}} = {\color{red}{\left(6 \int{\cos{\left(3 t \right)} d t}\right)}}$$

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 becomes

$$6 {\color{red}{\int{\cos{\left(3 t \right)} d t}}} = 6 {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}}$$

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

$$6 {\color{red}{\int{\frac{\cos{\left(u \right)}}{3} d u}}} = 6 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{3}\right)}}$$

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

$$2 {\color{red}{\int{\cos{\left(u \right)} d u}}} = 2 {\color{red}{\sin{\left(u \right)}}}$$

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

$$2 \sin{\left({\color{red}{u}} \right)} = 2 \sin{\left({\color{red}{\left(3 t\right)}} \right)}$$

Therefore,

$$\int{6 \cos{\left(3 t \right)} d t} = 2 \sin{\left(3 t \right)}$$

Add the constant of integration:

$$\int{6 \cos{\left(3 t \right)} d t} = 2 \sin{\left(3 t \right)}+C$$

$$$\int 6 \cos{\left(3 t \right)}\, dt = 2 \sin{\left(3 t \right)} + C$$$A