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

La calculadora encontrará la integral/antiderivada de $\cos^{2}{\left(2 x \right)}$, con los pasos que se muestran.

Escriba sin diferenciales como $dx$, $dy$ etc.
Deje vacío para la detección automática.

Si la calculadora no calculó algo o ha identificado un error, o tiene una sugerencia/comentario, escríbalo en los comentarios a continuación.

### Tu aportación

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

### Solución

Let $u=2 x$.

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

The integral becomes

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

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 $du = \frac{dv}{2}$.

Therefore,

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

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,

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

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

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