Integral of $$$\sin{\left(9 x \right)}$$$

Find $$$\int \sin{\left(9 x \right)}\, dx$$$.

Let $$$u=9 x$$$.

Then $$$du=\left(9 x\right)^{\prime }dx = 9 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{9}$$$.

The integral can be rewritten as

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{9}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Recall that $$$u=9 x$$$:

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

Therefore,

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

Add the constant of integration:

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

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