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

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

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

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

Let $$$u=\sin{\left(2 x \right)}$$$.

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

So,

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

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

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=1$$$:

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

Recall that $$$u=\sin{\left(2 x \right)}$$$:

$$\frac{{\color{red}{u}}^{2}}{8} = \frac{{\color{red}{\sin{\left(2 x \right)}}}^{2}}{8}$$

Therefore,

$$\int{\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2} d x} = \frac{\sin^{2}{\left(2 x \right)}}{8}$$

Add the constant of integration:

$$\int{\frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2} d x} = \frac{\sin^{2}{\left(2 x \right)}}{8}+C$$

$$$\int \frac{\sin{\left(2 x \right)} \cos{\left(2 x \right)}}{2}\, dx = \frac{\sin^{2}{\left(2 x \right)}}{8} + C$$$A