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

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

Apply the power reducing formula $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha=2 x$$$:

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

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 $$$f{\left(x \right)} = \frac{\left(\cos{\left(4 x \right)} + 1\right) \sin{\left(2 x \right)}}{2}$$$:

$${\color{red}{\int{\frac{\left(\cos{\left(4 x \right)} + 1\right) \sin{\left(2 x \right)}}{4} d x}}} = {\color{red}{\left(\frac{\int{\frac{\left(\cos{\left(4 x \right)} + 1\right) \sin{\left(2 x \right)}}{2} d x}}{2}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\frac{\left(\cos{\left(4 x \right)} + 1\right) \sin{\left(2 x \right)}}{2} d x}}}}{2} = \frac{{\color{red}{\int{\left(\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{2}\right)d x}}}}{2}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{2}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x} + \int{\frac{\sin{\left(2 x \right)}}{2} d x}\right)}}}{2}$$

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

$$\frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}}{2} = \frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}}{2}$$

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

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

The integral can be rewritten as

$$\frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{4} = \frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4}$$

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

$$\frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{4} = \frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{4}$$

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

$$\frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = \frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

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

$$\frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}{2} - \frac{\cos{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

Rewrite $$$\sin\left(2 x \right)\cos\left(4 x \right)$$$ 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 $$$\alpha=2 x$$$ and $$$\beta=4 x$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)} \cos{\left(4 x \right)}}{2} d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(6 x \right)}}{4}\right)d x}}}}{2}$$

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 $$$f{\left(x \right)} = - \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{4} + \frac{\sin{\left(6 x \right)}}{4}\right)d x}}}}{2} = - \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\left(\frac{\int{\left(- \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}\right)d x}}{2}\right)}}}{2}$$

Integrate term by term:

$$- \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}\right)d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} + \frac{{\color{red}{\left(- \int{\frac{\sin{\left(2 x \right)}}{2} d x} + \int{\frac{\sin{\left(6 x \right)}}{2} d x}\right)}}}{4}$$

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 $$$f{\left(x \right)} = \sin{\left(6 x \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(6 x \right)}}{2} d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(6 x \right)} d x}}{2}\right)}}}{4}$$

Let $$$u=6 x$$$.

Then $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{6}$$$.

The integral can be rewritten as

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(6 x \right)} d x}}}}{8} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{8}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{6}$$$ and $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{8} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{6}\right)}}}{8}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{48}$$

Recall that $$$u=6 x$$$:

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{u}} \right)}}{48} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}{4} - \frac{\cos{\left({\color{red}{\left(6 x\right)}} \right)}}{48}$$

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

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\frac{\sin{\left(2 x \right)}}{2} d x}}}}{4} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(2 x \right)} d x}}{2}\right)}}}{4}$$

The integral $$$\int{\sin{\left(2 x \right)} d x}$$$ was already calculated:

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

Therefore,

$$- \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{8} = - \frac{\cos{\left(2 x \right)}}{8} - \frac{\cos{\left(6 x \right)}}{48} - \frac{{\color{red}{\left(- \frac{\cos{\left(2 x \right)}}{2}\right)}}}{8}$$

Therefore,

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2} d x} = - \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}+C$$

$$$\int \frac{\sin{\left(2 x \right)} \cos^{2}{\left(2 x \right)}}{2}\, dx = \left(- \frac{\cos{\left(2 x \right)}}{16} - \frac{\cos{\left(6 x \right)}}{48}\right) + C$$$A