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

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

Rewrite the integrand 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 $$$\alpha=2 x$$$ and $$$\beta=x$$$:

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

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

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

Integrate term by term:

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

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

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

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

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

So,

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

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

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

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

$$- \frac{\cos{\left(x \right)}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{6} = - \frac{\cos{\left(x \right)}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{6}$$

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

$$- \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{6} = - \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{6}$$

Therefore,

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{6}$$

Add the constant of integration:

$$\int{\sin{\left(2 x \right)} \cos{\left(x \right)} d x} = - \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{6}+C$$

$$$\int \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx = \left(- \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{6}\right) + C$$$A