Integral de $$$- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}$$$

Halla $$$\int \left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)\, dt$$$.

Integra término a término:

$${\color{red}{\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t}}} = {\color{red}{\left(\int{\sin^{2}{\left(t \right)} d t} - \int{\sin^{4}{\left(t \right)} d t}\right)}}$$

Aplica la fórmula de reducción de potencia $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ con $$$\alpha=t$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(t \right)} = 1 - \cos{\left(2 t \right)}$$$:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dt = c t$$$ con $$$c=1$$$:

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

Sea $$$u=2 t$$$.

Entonces $$$du=\left(2 t\right)^{\prime }dt = 2 dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = \frac{du}{2}$$$.

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Recordemos que $$$u=2 t$$$:

$$\frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{t}{2} - \int{\sin^{4}{\left(t \right)} d t} - \frac{\sin{\left({\color{red}{\left(2 t\right)}} \right)}}{4}$$

Aplica la fórmula de reducción de potencia $$$\sin^{4}{\left(\alpha \right)} = - \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{\cos{\left(4 \alpha \right)}}{8} + \frac{3}{8}$$$ con $$$\alpha=t$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(t \right)} = - 4 \cos{\left(2 t \right)} + \cos{\left(4 t \right)} + 3$$$:

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dt = c t$$$ con $$$c=3$$$:

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

Aplica la regla del factor constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ con $$$c=4$$$ y $$$f{\left(t \right)} = \cos{\left(2 t \right)}$$$:

$$\frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\int{4 \cos{\left(2 t \right)} d t}}}}{8} = \frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\left(4 \int{\cos{\left(2 t \right)} d t}\right)}}}{8}$$

La integral $$$\int{\cos{\left(2 t \right)} d t}$$$ ya ha sido calculada:

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

Por lo tanto,

$$\frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\int{\cos{\left(2 t \right)} d t}}}}{2} = \frac{t}{8} - \frac{\sin{\left(2 t \right)}}{4} - \frac{\int{\cos{\left(4 t \right)} d t}}{8} + \frac{{\color{red}{\left(\frac{\sin{\left(2 t \right)}}{2}\right)}}}{2}$$

Sea $$$v=4 t$$$.

Entonces $$$dv=\left(4 t\right)^{\prime }dt = 4 dt$$$ (los pasos pueden verse »), y obtenemos que $$$dt = \frac{dv}{4}$$$.

Entonces,

$$\frac{t}{8} - \frac{{\color{red}{\int{\cos{\left(4 t \right)} d t}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{8}$$

Aplica la regla del factor constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ con $$$c=\frac{1}{4}$$$ y $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{t}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{8} = \frac{t}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{8}$$

La integral del coseno es $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{t}{8} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{32} = \frac{t}{8} - \frac{{\color{red}{\sin{\left(v \right)}}}}{32}$$

Recordemos que $$$v=4 t$$$:

$$\frac{t}{8} - \frac{\sin{\left({\color{red}{v}} \right)}}{32} = \frac{t}{8} - \frac{\sin{\left({\color{red}{\left(4 t\right)}} \right)}}{32}$$

Por lo tanto,

$$\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}$$

Añade la constante de integración:

$$\int{\left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)d t} = \frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}+C$$

$$$\int \left(- \sin^{4}{\left(t \right)} + \sin^{2}{\left(t \right)}\right)\, dt = \left(\frac{t}{8} - \frac{\sin{\left(4 t \right)}}{32}\right) + C$$$A