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

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

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 here), and we have that $$$\cos{\left(2 x \right)} dx = \frac{du}{2}$$$.

Therefore,

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

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

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

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

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

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

$$\frac{\color{red}{u}^{2}}{4} = \frac{\color{red}{\sin{\left(2 x \right)}}^{2}}{4}$$

Therefore,

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

Add the constant of integration:

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

Answer: $$$\int{\sin{\left(2 x \right)} \cos{\left(2 x \right)} d x}=\frac{\sin^{2}{\left(2 x \right)}}{4}+C$$$