Integral of $$$4 \sin{\left(6 x \right)}$$$

Find $$$\int 4 \sin{\left(6 x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \sin{\left(6 x \right)}$$$:

$${\color{red}{\int{4 \sin{\left(6 x \right)} d x}}} = {\color{red}{\left(4 \int{\sin{\left(6 x \right)} d x}\right)}}$$

Let $$$u=6 x$$$.

Then $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{6}$$$.

So,

$$4 {\color{red}{\int{\sin{\left(6 x \right)} d x}}} = 4 {\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$4 {\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}} = 4 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{6}\right)}}$$

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

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

Recall that $$$u=6 x$$$:

$$- \frac{2 \cos{\left({\color{red}{u}} \right)}}{3} = - \frac{2 \cos{\left({\color{red}{\left(6 x\right)}} \right)}}{3}$$

Therefore,

$$\int{4 \sin{\left(6 x \right)} d x} = - \frac{2 \cos{\left(6 x \right)}}{3}$$

Add the constant of integration:

$$\int{4 \sin{\left(6 x \right)} d x} = - \frac{2 \cos{\left(6 x \right)}}{3}+C$$

$$$\int 4 \sin{\left(6 x \right)}\, dx = - \frac{2 \cos{\left(6 x \right)}}{3} + C$$$A