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

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

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

$${\color{red}{\int{9 x \cos{\left(x \right)} d x}}} = {\color{red}{\left(9 \int{x \cos{\left(x \right)} d x}\right)}}$$

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

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

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

The integral becomes

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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