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

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

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 $$$dx = 3 du$$$.

Thus,

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

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

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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