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

A calculadora encontrará a integral/antiderivada de $\cos^{2}{\left(2 x \right)}$, com as etapas mostradas.

Por favor, escreva sem nenhum diferencial como $dx$, $dy$ etc.
Deixe em branco para detecção automática.

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

### Solução

Let $u=2 x$.

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

Therefore,

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

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

Thus,

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

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