Integral de $$$\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}$$$

Encontre $$$\int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$$.

Reescreva o integrando usando as fórmulas de redução de potência $$$\sin^2\left( \alpha \right)=\frac{1}{2}-\frac{1}{2}\cos\left(2 \alpha \right)-$$$ com $$$\alpha=x$$$ e $$$\cos^2\left( \beta \right)=\frac{1}{2}+\frac{1}{2}\cos\left(2 \beta \right)+$$$ com $$$\beta=x$$$:

$${\color{red}{\int{\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) d x}}}$$

Expanda a expressão:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)^{2} \left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right) d x}}} = {\color{red}{\int{\left(\frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}\right)d x}}}$$

Integre termo a termo:

$${\color{red}{\int{\left(\frac{\cos^{3}{\left(2 x \right)}}{8} - \frac{\cos^{2}{\left(2 x \right)}}{8} - \frac{\cos{\left(2 x \right)}}{8} + \frac{1}{8}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{8} d x} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x}\right)}}$$

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=\frac{1}{8}$$$:

$$- \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} + {\color{red}{\int{\frac{1}{8} d x}}} = - \int{\frac{\cos{\left(2 x \right)}}{8} d x} - \int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} + {\color{red}{\left(\frac{x}{8}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{8}$$$ e $$$f{\left(x \right)} = \cos^{2}{\left(2 x \right)}$$$:

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\int{\frac{\cos^{2}{\left(2 x \right)}}{8} d x}}} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\left(\frac{\int{\cos^{2}{\left(2 x \right)} d x}}{8}\right)}}$$

Seja $$$u=2 x$$$.

Então $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.

Portanto,

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{2}{\left(2 x \right)} d x}}}}{8} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8}$$

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

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{2} d u}}}}{8} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\cos^{2}{\left(u \right)} d u}}{2}\right)}}}{8}$$

Aplique a fórmula de redução de potência $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ com $$$\alpha= u $$$:

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{16} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16}$$

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

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{16} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{16}$$

Integre termo a termo:

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{32} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\int{1 d u} + \int{\cos{\left(2 u \right)} d u}\right)}}}{32}$$

Aplique a regra da constante $$$\int c\, du = c u$$$ usando $$$c=1$$$:

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{\int{1 d u}}}}{32} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\int{\cos{\left(2 u \right)} d u}}{32} - \frac{{\color{red}{u}}}{32}$$

Seja $$$v=2 u$$$.

Então $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (veja os passos »), e obtemos $$$du = \frac{dv}{2}$$$.

A integral pode ser reescrita como

$$- \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{32} = - \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$- \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{32} = - \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{32}$$

A integral do cosseno é $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$- \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{64} = - \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\sin{\left(v \right)}}}}{64}$$

Recorde que $$$v=2 u$$$:

$$- \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{v}} \right)}}{64} = - \frac{u}{32} + \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{64}$$

Recorde que $$$u=2 x$$$:

$$\frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left(2 {\color{red}{u}} \right)}}{64} - \frac{{\color{red}{u}}}{32} = \frac{x}{8} - \int{\frac{\cos{\left(2 x \right)}}{8} d x} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{\sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{64} - \frac{{\color{red}{\left(2 x\right)}}}{32}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{8}$$$ e $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\int{\frac{\cos{\left(2 x \right)}}{8} d x}}} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - {\color{red}{\left(\frac{\int{\cos{\left(2 x \right)} d x}}{8}\right)}}$$

A integral $$$\int{\cos{\left(2 x \right)} d x}$$$ já foi calculada:

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

Portanto,

$$\frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{8} = \frac{x}{16} - \frac{\sin{\left(4 x \right)}}{64} + \int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{8}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{1}{8}$$$ e $$$f{\left(x \right)} = \cos^{3}{\left(2 x \right)}$$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + {\color{red}{\int{\frac{\cos^{3}{\left(2 x \right)}}{8} d x}}} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + {\color{red}{\left(\frac{\int{\cos^{3}{\left(2 x \right)} d x}}{8}\right)}}$$

Seja $$$v=2 x$$$.

Então $$$dv=\left(2 x\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{dv}{2}$$$.

A integral torna-se

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\cos^{3}{\left(2 x \right)} d x}}}}{8} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos^{3}{\left(v \right)}}{2} d v}}}}{8}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(v \right)} = \cos^{3}{\left(v \right)}$$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\frac{\cos^{3}{\left(v \right)}}{2} d v}}}}{8} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\left(\frac{\int{\cos^{3}{\left(v \right)} d v}}{2}\right)}}}{8}$$

Separe um cosseno e escreva o restante em termos de seno, usando a fórmula $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ com $$$\alpha= v $$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\cos^{3}{\left(v \right)} d v}}}}{16} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(v \right)}\right) \cos{\left(v \right)} d v}}}}{16}$$

Seja $$$w=\sin{\left(v \right)}$$$.

Então $$$dw=\left(\sin{\left(v \right)}\right)^{\prime }dv = \cos{\left(v \right)} dv$$$ (veja os passos »), e obtemos $$$\cos{\left(v \right)} dv = dw$$$.

Assim,

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\left(1 - \sin^{2}{\left(v \right)}\right) \cos{\left(v \right)} d v}}}}{16} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\left(1 - w^{2}\right)d w}}}}{16}$$

Integre termo a termo:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\int{\left(1 - w^{2}\right)d w}}}}{16} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\left(\int{1 d w} - \int{w^{2} d w}\right)}}}{16}$$

Aplique a regra da constante $$$\int c\, dw = c w$$$ usando $$$c=1$$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} - \frac{\int{w^{2} d w}}{16} + \frac{{\color{red}{\int{1 d w}}}}{16} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} - \frac{\int{w^{2} d w}}{16} + \frac{{\color{red}{w}}}{16}$$

Aplique a regra da potência $$$\int w^{n}\, dw = \frac{w^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=2$$$:

$$\frac{w}{16} + \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} - \frac{{\color{red}{\int{w^{2} d w}}}}{16}=\frac{w}{16} + \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} - \frac{{\color{red}{\frac{w^{1 + 2}}{1 + 2}}}}{16}=\frac{w}{16} + \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} - \frac{{\color{red}{\left(\frac{w^{3}}{3}\right)}}}{16}$$

Recorde que $$$w=\sin{\left(v \right)}$$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{w}}}{16} - \frac{{\color{red}{w}}^{3}}{48} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{{\color{red}{\sin{\left(v \right)}}}}{16} - \frac{{\color{red}{\sin{\left(v \right)}}}^{3}}{48}$$

Recorde que $$$v=2 x$$$:

$$\frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{\sin{\left({\color{red}{v}} \right)}}{16} - \frac{\sin^{3}{\left({\color{red}{v}} \right)}}{48} = \frac{x}{16} - \frac{\sin{\left(2 x \right)}}{16} - \frac{\sin{\left(4 x \right)}}{64} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{16} - \frac{\sin^{3}{\left({\color{red}{\left(2 x\right)}} \right)}}{48}$$

Portanto,

$$\int{\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}$$

Adicione a constante de integração:

$$\int{\sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}+C$$

$$$\int \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = \left(\frac{x}{16} - \frac{\sin^{3}{\left(2 x \right)}}{48} - \frac{\sin{\left(4 x \right)}}{64}\right) + C$$$A