Integral of $$$\frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}}$$$ with respect to $$$x$$$

Find $$$\int \frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{s^{2} \sin{\left(1 \right)}}{c_{0}}$$$ and $$$f{\left(x \right)} = x \cos{\left(2 x \right)}$$$:

$${\color{red}{\int{\frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}} d x}}} = {\color{red}{\frac{s^{2} \sin{\left(1 \right)} \int{x \cos{\left(2 x \right)} d x}}{c_{0}}}}$$

For the integral $$$\int{x \cos{\left(2 x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=\cos{\left(2 x \right)} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{\cos{\left(2 x \right)} d x}=\frac{\sin{\left(2 x \right)}}{2}$$$ (steps can be seen »).

Thus,

$$\frac{s^{2} \sin{\left(1 \right)} {\color{red}{\int{x \cos{\left(2 x \right)} d x}}}}{c_{0}}=\frac{s^{2} \sin{\left(1 \right)} {\color{red}{\left(x \cdot \frac{\sin{\left(2 x \right)}}{2}-\int{\frac{\sin{\left(2 x \right)}}{2} \cdot 1 d x}\right)}}}{c_{0}}=\frac{s^{2} \sin{\left(1 \right)} {\color{red}{\left(\frac{x \sin{\left(2 x \right)}}{2} - \int{\frac{\sin{\left(2 x \right)}}{2} d x}\right)}}}{c_{0}}$$

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 $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$$\frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}\right)}{c_{0}} = \frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - {\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}\right)}{c_{0}}$$

Let $$$u=2 x$$$.

Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Therefore,

$$\frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{2}\right)}{c_{0}} = \frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}\right)}{c_{0}}$$

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

$$\frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{2}\right)}{c_{0}} = \frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{2}\right)}{c_{0}}$$

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

$$\frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{4}\right)}{c_{0}} = \frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{4}\right)}{c_{0}}$$

Recall that $$$u=2 x$$$:

$$\frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{u}} \right)}}{4}\right)}{c_{0}} = \frac{s^{2} \sin{\left(1 \right)} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{4}\right)}{c_{0}}$$

Therefore,

$$\int{\frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}} d x} = \frac{s^{2} \left(\frac{x \sin{\left(2 x \right)}}{2} + \frac{\cos{\left(2 x \right)}}{4}\right) \sin{\left(1 \right)}}{c_{0}}$$

Simplify:

$$\int{\frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}} d x} = \frac{s^{2} \left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) \sin{\left(1 \right)}}{4 c_{0}}$$

Add the constant of integration:

$$\int{\frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}} d x} = \frac{s^{2} \left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) \sin{\left(1 \right)}}{4 c_{0}}+C$$

$$$\int \frac{s^{2} x \sin{\left(1 \right)} \cos{\left(2 x \right)}}{c_{0}}\, dx = \frac{s^{2} \left(2 x \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) \sin{\left(1 \right)}}{4 c_{0}} + C$$$A