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

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

Rewrite $$$\sin\left(x \right)\cos\left(2 x \right)$$$ using the formula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ with $$$\alpha=x$$$ and $$$\beta=2 x$$$:

$${\color{red}{\int{\left(- 2 \sin{\left(x \right)} \cos{\left(2 x \right)}\right)d x}}} = {\color{red}{\int{\left(\sin{\left(x \right)} - \sin{\left(3 x \right)}\right)d x}}}$$

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 $$$f{\left(x \right)} = 2 \sin{\left(x \right)} - 2 \sin{\left(3 x \right)}$$$:

$${\color{red}{\int{\left(\sin{\left(x \right)} - \sin{\left(3 x \right)}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(2 \sin{\left(x \right)} - 2 \sin{\left(3 x \right)}\right)d x}}{2}\right)}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(2 \sin{\left(x \right)} - 2 \sin{\left(3 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{2 \sin{\left(x \right)} d x} - \int{2 \sin{\left(3 x \right)} d x}\right)}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$:

$$\frac{\int{2 \sin{\left(x \right)} d x}}{2} - \frac{{\color{red}{\int{2 \sin{\left(3 x \right)} d x}}}}{2} = \frac{\int{2 \sin{\left(x \right)} d x}}{2} - \frac{{\color{red}{\left(2 \int{\sin{\left(3 x \right)} d x}\right)}}}{2}$$

Let $$$u=3 x$$$.

Then $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

Thus,

$$\frac{\int{2 \sin{\left(x \right)} d x}}{2} - {\color{red}{\int{\sin{\left(3 x \right)} d x}}} = \frac{\int{2 \sin{\left(x \right)} d x}}{2} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\int{2 \sin{\left(x \right)} d x}}{2} - {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}} = \frac{\int{2 \sin{\left(x \right)} d x}}{2} - {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}$$

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

$$\frac{\int{2 \sin{\left(x \right)} d x}}{2} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{3} = \frac{\int{2 \sin{\left(x \right)} d x}}{2} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{3}$$

Recall that $$$u=3 x$$$:

$$\frac{\int{2 \sin{\left(x \right)} d x}}{2} + \frac{\cos{\left({\color{red}{u}} \right)}}{3} = \frac{\int{2 \sin{\left(x \right)} d x}}{2} + \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{3}$$

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

$$\frac{\cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\int{2 \sin{\left(x \right)} d x}}}}{2} = \frac{\cos{\left(3 x \right)}}{3} + \frac{{\color{red}{\left(2 \int{\sin{\left(x \right)} d x}\right)}}}{2}$$

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

$$\frac{\cos{\left(3 x \right)}}{3} + {\color{red}{\int{\sin{\left(x \right)} d x}}} = \frac{\cos{\left(3 x \right)}}{3} + {\color{red}{\left(- \cos{\left(x \right)}\right)}}$$

Therefore,

$$\int{\left(- 2 \sin{\left(x \right)} \cos{\left(2 x \right)}\right)d x} = - \cos{\left(x \right)} + \frac{\cos{\left(3 x \right)}}{3}$$

Add the constant of integration:

$$\int{\left(- 2 \sin{\left(x \right)} \cos{\left(2 x \right)}\right)d x} = - \cos{\left(x \right)} + \frac{\cos{\left(3 x \right)}}{3}+C$$

$$$\int \left(- 2 \sin{\left(x \right)} \cos{\left(2 x \right)}\right)\, dx = \left(- \cos{\left(x \right)} + \frac{\cos{\left(3 x \right)}}{3}\right) + C$$$A