Integral de $$$\cos^{2}{\left(2 x \right)}$$$

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

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

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

So,

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

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 $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$:

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

Rewrite the cosine using the power reducing formula $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha= u $$$:

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

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 $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Let $$$v=2 u$$$.

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

The integral can be rewritten as

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

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

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

The integral of the cosine is $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

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

Recall that $$$v=2 u$$$:

$$\frac{u}{4} + \frac{\sin{\left({\color{red}{v}} \right)}}{8} = \frac{u}{4} + \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{8}$$

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

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

Therefore,

$$\int{\cos^{2}{\left(2 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}$$

Add the constant of integration:

$$\int{\cos^{2}{\left(2 x \right)} d x} = \frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}+C$$

Answer: $$$\int{\cos^{2}{\left(2 x \right)} d x}=\frac{x}{2} + \frac{\sin{\left(4 x \right)}}{8}+C$$$