Integral de $$$6 \cos^{2}{\left(x \right)}$$$
Calculadora relacionada: Calculadora de Integrais Definidas e Impróprias
Sua entrada
Encontre $$$\int 6 \cos^{2}{\left(x \right)}\, dx$$$.
Solução
Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=6$$$ e $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$:
$${\color{red}{\int{6 \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\left(6 \int{\cos^{2}{\left(x \right)} d x}\right)}}$$
Aplique a fórmula de redução de potência $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ com $$$\alpha=x$$$:
$$6 {\color{red}{\int{\cos^{2}{\left(x \right)} d x}}} = 6 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$:
$$6 {\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}} = 6 {\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}$$
Integre termo a termo:
$$3 {\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}} = 3 {\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}$$
Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=1$$$:
$$3 \int{\cos{\left(2 x \right)} d x} + 3 {\color{red}{\int{1 d x}}} = 3 \int{\cos{\left(2 x \right)} d x} + 3 {\color{red}{x}}$$
Seja $$$u=2 x$$$.
Então $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.
A integral pode ser reescrita como
$$3 x + 3 {\color{red}{\int{\cos{\left(2 x \right)} d x}}} = 3 x + 3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}$$
Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$3 x + 3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}} = 3 x + 3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}$$
A integral do cosseno é $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$3 x + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{2} = 3 x + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{2}$$
Recorde que $$$u=2 x$$$:
$$3 x + \frac{3 \sin{\left({\color{red}{u}} \right)}}{2} = 3 x + \frac{3 \sin{\left({\color{red}{\left(2 x\right)}} \right)}}{2}$$
Portanto,
$$\int{6 \cos^{2}{\left(x \right)} d x} = 3 x + \frac{3 \sin{\left(2 x \right)}}{2}$$
Adicione a constante de integração:
$$\int{6 \cos^{2}{\left(x \right)} d x} = 3 x + \frac{3 \sin{\left(2 x \right)}}{2}+C$$
Resposta
$$$\int 6 \cos^{2}{\left(x \right)}\, dx = \left(3 x + \frac{3 \sin{\left(2 x \right)}}{2}\right) + C$$$A