Integral of $$$- 3 \cos{\left(\frac{x}{3} \right)}$$$

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

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

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

Let $$$u=\frac{x}{3}$$$.

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

Therefore,

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

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

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

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

$$- 9 {\color{red}{\int{\cos{\left(u \right)} d u}}} = - 9 {\color{red}{\sin{\left(u \right)}}}$$

Recall that $$$u=\frac{x}{3}$$$:

$$- 9 \sin{\left({\color{red}{u}} \right)} = - 9 \sin{\left({\color{red}{\left(\frac{x}{3}\right)}} \right)}$$

Therefore,

$$\int{\left(- 3 \cos{\left(\frac{x}{3} \right)}\right)d x} = - 9 \sin{\left(\frac{x}{3} \right)}$$

Add the constant of integration:

$$\int{\left(- 3 \cos{\left(\frac{x}{3} \right)}\right)d x} = - 9 \sin{\left(\frac{x}{3} \right)}+C$$

$$$\int \left(- 3 \cos{\left(\frac{x}{3} \right)}\right)\, dx = - 9 \sin{\left(\frac{x}{3} \right)} + C$$$A