Integral of $$$\frac{x \cos{\left(x \right)}}{2}$$$

Find $$$\int \frac{x \cos{\left(x \right)}}{2}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = x \cos{\left(x \right)}$$$:

$${\color{red}{\int{\frac{x \cos{\left(x \right)}}{2} d x}}} = {\color{red}{\left(\frac{\int{x \cos{\left(x \right)} d x}}{2}\right)}}$$

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 »).

So,

$$\frac{{\color{red}{\int{x \cos{\left(x \right)} d x}}}}{2}=\frac{{\color{red}{\left(x \cdot \sin{\left(x \right)}-\int{\sin{\left(x \right)} \cdot 1 d x}\right)}}}{2}=\frac{{\color{red}{\left(x \sin{\left(x \right)} - \int{\sin{\left(x \right)} d x}\right)}}}{2}$$

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

$$\frac{x \sin{\left(x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{2} = \frac{x \sin{\left(x \right)}}{2} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{2}$$

Therefore,

$$\int{\frac{x \cos{\left(x \right)}}{2} d x} = \frac{x \sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}$$

Simplify:

$$\int{\frac{x \cos{\left(x \right)}}{2} d x} = \frac{x \sin{\left(x \right)} + \cos{\left(x \right)}}{2}$$

Add the constant of integration:

$$\int{\frac{x \cos{\left(x \right)}}{2} d x} = \frac{x \sin{\left(x \right)} + \cos{\left(x \right)}}{2}+C$$

$$$\int \frac{x \cos{\left(x \right)}}{2}\, dx = \frac{x \sin{\left(x \right)} + \cos{\left(x \right)}}{2} + C$$$A