Integral of $$$6 \sin{\left(3 x \right)}$$$

Find $$$\int 6 \sin{\left(3 x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=6$$$ and $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$:

$${\color{red}{\int{6 \sin{\left(3 x \right)} d x}}} = {\color{red}{\left(6 \int{\sin{\left(3 x \right)} d x}\right)}}$$

Let $$$u=3 x$$$.

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

So,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int 6 \sin{\left(3 x \right)}\, dx = - 2 \cos{\left(3 x \right)} + C$$$A