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

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

For the integral $$$\int{x \cos{\left(9 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=\cos{\left(9 x \right)} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(9 x \right)} d x}=\frac{\sin{\left(9 x \right)}}{9}$$$ (steps can be seen »).

The integral can be rewritten as

$${\color{red}{\int{x \cos{\left(9 x \right)} d x}}}={\color{red}{\left(x \cdot \frac{\sin{\left(9 x \right)}}{9}-\int{\frac{\sin{\left(9 x \right)}}{9} \cdot 1 d x}\right)}}={\color{red}{\left(\frac{x \sin{\left(9 x \right)}}{9} - \int{\frac{\sin{\left(9 x \right)}}{9} d x}\right)}}$$

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

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

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

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

So,

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

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

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

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

$$\frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{81} = \frac{x \sin{\left(9 x \right)}}{9} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{81}$$

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

$$\frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left({\color{red}{u}} \right)}}{81} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left({\color{red}{\left(9 x\right)}} \right)}}{81}$$

Therefore,

$$\int{x \cos{\left(9 x \right)} d x} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}$$

Add the constant of integration:

$$\int{x \cos{\left(9 x \right)} d x} = \frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}+C$$

$$$\int x \cos{\left(9 x \right)}\, dx = \left(\frac{x \sin{\left(9 x \right)}}{9} + \frac{\cos{\left(9 x \right)}}{81}\right) + C$$$A