Integral de $$$\sin{\left(5 x \right)} \cos^{5}{\left(x \right)}$$$

Halla $$$\int \sin{\left(5 x \right)} \cos^{5}{\left(x \right)}\, dx$$$.

Aplica la fórmula de reducción de potencia $$$\cos^{5}{\left(\alpha \right)} = \frac{5 \cos{\left(\alpha \right)}}{8} + \frac{5 \cos{\left(3 \alpha \right)}}{16} + \frac{\cos{\left(5 \alpha \right)}}{16}$$$ con $$$\alpha=x$$$:

$${\color{red}{\int{\sin{\left(5 x \right)} \cos^{5}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{\left(10 \cos{\left(x \right)} + 5 \cos{\left(3 x \right)} + \cos{\left(5 x \right)}\right) \sin{\left(5 x \right)}}{16} d x}}}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{16}$$$ y $$$f{\left(x \right)} = \left(10 \cos{\left(x \right)} + 5 \cos{\left(3 x \right)} + \cos{\left(5 x \right)}\right) \sin{\left(5 x \right)}$$$:

$${\color{red}{\int{\frac{\left(10 \cos{\left(x \right)} + 5 \cos{\left(3 x \right)} + \cos{\left(5 x \right)}\right) \sin{\left(5 x \right)}}{16} d x}}} = {\color{red}{\left(\frac{\int{\left(10 \cos{\left(x \right)} + 5 \cos{\left(3 x \right)} + \cos{\left(5 x \right)}\right) \sin{\left(5 x \right)} d x}}{16}\right)}}$$

Expand the expression:

$$\frac{{\color{red}{\int{\left(10 \cos{\left(x \right)} + 5 \cos{\left(3 x \right)} + \cos{\left(5 x \right)}\right) \sin{\left(5 x \right)} d x}}}}{16} = \frac{{\color{red}{\int{\left(10 \sin{\left(5 x \right)} \cos{\left(x \right)} + 5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} + \sin{\left(5 x \right)} \cos{\left(5 x \right)}\right)d x}}}}{16}$$

Integra término a término:

$$\frac{{\color{red}{\int{\left(10 \sin{\left(5 x \right)} \cos{\left(x \right)} + 5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} + \sin{\left(5 x \right)} \cos{\left(5 x \right)}\right)d x}}}}{16} = \frac{{\color{red}{\left(\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x} + \int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x} + \int{\sin{\left(5 x \right)} \cos{\left(5 x \right)} d x}\right)}}}{16}$$

Sea $$$u=\sin{\left(5 x \right)}$$$.

Entonces $$$du=\left(\sin{\left(5 x \right)}\right)^{\prime }dx = 5 \cos{\left(5 x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\cos{\left(5 x \right)} dx = \frac{du}{5}$$$.

La integral puede reescribirse como

$$\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\int{\sin{\left(5 x \right)} \cos{\left(5 x \right)} d x}}}}{16} = \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\int{\frac{u}{5} d u}}}}{16}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{5}$$$ y $$$f{\left(u \right)} = u$$$:

$$\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\int{\frac{u}{5} d u}}}}{16} = \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\left(\frac{\int{u d u}}{5}\right)}}}{16}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:

$$\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\int{u d u}}}}{80}=\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\frac{u^{1 + 1}}{1 + 1}}}}{80}=\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\left(\frac{u^{2}}{2}\right)}}}{80}$$

Recordemos que $$$u=\sin{\left(5 x \right)}$$$:

$$\frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{u}}^{2}}{160} = \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}{16} + \frac{{\color{red}{\sin{\left(5 x \right)}}}^{2}}{160}$$

Reescribe $$$\sin\left(5 x \right)\cos\left(3 x \right)$$$ utilizando la fórmula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ con $$$\alpha=5 x$$$ y $$$\beta=3 x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\int{5 \sin{\left(5 x \right)} \cos{\left(3 x \right)} d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\int{\left(\frac{5 \sin{\left(2 x \right)}}{2} + \frac{5 \sin{\left(8 x \right)}}{2}\right)d x}}}}{16}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(x \right)} = 5 \sin{\left(2 x \right)} + 5 \sin{\left(8 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\int{\left(\frac{5 \sin{\left(2 x \right)}}{2} + \frac{5 \sin{\left(8 x \right)}}{2}\right)d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\left(\frac{\int{\left(5 \sin{\left(2 x \right)} + 5 \sin{\left(8 x \right)}\right)d x}}{2}\right)}}}{16}$$

Integra término a término:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\int{\left(5 \sin{\left(2 x \right)} + 5 \sin{\left(8 x \right)}\right)d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\left(\int{5 \sin{\left(2 x \right)} d x} + \int{5 \sin{\left(8 x \right)} d x}\right)}}}{32}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=5$$$ y $$$f{\left(x \right)} = \sin{\left(2 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{{\color{red}{\int{5 \sin{\left(2 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{{\color{red}{\left(5 \int{\sin{\left(2 x \right)} d x}\right)}}}{32}$$

Sea $$$u=2 x$$$.

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

Entonces,

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\sin{\left(2 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{32}$$

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 $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}}{32}$$

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

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{64} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} + \frac{5 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{64}$$

Recordemos que $$$u=2 x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} - \frac{5 \cos{\left({\color{red}{u}} \right)}}{64} = \frac{\sin^{2}{\left(5 x \right)}}{160} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{\int{5 \sin{\left(8 x \right)} d x}}{32} - \frac{5 \cos{\left({\color{red}{\left(2 x\right)}} \right)}}{64}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=5$$$ y $$$f{\left(x \right)} = \sin{\left(8 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\int{5 \sin{\left(8 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{{\color{red}{\left(5 \int{\sin{\left(8 x \right)} d x}\right)}}}{32}$$

Sea $$$u=8 x$$$.

Entonces $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{8}$$$.

La integral se convierte en

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\int{\sin{\left(8 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{8} d u}}}}{32}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{8}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{8} d u}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{8}\right)}}}{32}$$

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

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{256} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} + \frac{5 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{256}$$

Recordemos que $$$u=8 x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} - \frac{5 \cos{\left({\color{red}{u}} \right)}}{256} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} + \frac{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}{16} - \frac{5 \cos{\left({\color{red}{\left(8 x\right)}} \right)}}{256}$$

Reescribe $$$\sin\left(5 x \right)\cos\left(x \right)$$$ utilizando la fórmula $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ con $$$\alpha=5 x$$$ y $$$\beta=x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\int{10 \sin{\left(5 x \right)} \cos{\left(x \right)} d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\int{\left(5 \sin{\left(4 x \right)} + 5 \sin{\left(6 x \right)}\right)d x}}}}{16}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{1}{2}$$$ y $$$f{\left(x \right)} = 10 \sin{\left(4 x \right)} + 10 \sin{\left(6 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\int{\left(5 \sin{\left(4 x \right)} + 5 \sin{\left(6 x \right)}\right)d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\left(\frac{\int{\left(10 \sin{\left(4 x \right)} + 10 \sin{\left(6 x \right)}\right)d x}}{2}\right)}}}{16}$$

Integra término a término:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\int{\left(10 \sin{\left(4 x \right)} + 10 \sin{\left(6 x \right)}\right)d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\left(\int{10 \sin{\left(4 x \right)} d x} + \int{10 \sin{\left(6 x \right)} d x}\right)}}}{32}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=10$$$ y $$$f{\left(x \right)} = \sin{\left(4 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\int{10 \sin{\left(4 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{{\color{red}{\left(10 \int{\sin{\left(4 x \right)} d x}\right)}}}{32}$$

Sea $$$u=4 x$$$.

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

Por lo tanto,

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\sin{\left(4 x \right)} d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{16}$$

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

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{4}\right)}}}{16}$$

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

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{64} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} + \frac{5 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{64}$$

Recordemos que $$$u=4 x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} - \frac{5 \cos{\left({\color{red}{u}} \right)}}{64} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{\int{10 \sin{\left(6 x \right)} d x}}{32} - \frac{5 \cos{\left({\color{red}{\left(4 x\right)}} \right)}}{64}$$

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=10$$$ y $$$f{\left(x \right)} = \sin{\left(6 x \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\int{10 \sin{\left(6 x \right)} d x}}}}{32} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{{\color{red}{\left(10 \int{\sin{\left(6 x \right)} d x}\right)}}}{32}$$

Sea $$$u=6 x$$$.

Entonces $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (los pasos pueden verse »), y obtenemos que $$$dx = \frac{du}{6}$$$.

Entonces,

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\int{\sin{\left(6 x \right)} d x}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{16}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{6}$$$ y $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\int{\frac{\sin{\left(u \right)}}{6} d u}}}}{16} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{6}\right)}}}{16}$$

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

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{96} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} + \frac{5 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{96}$$

Recordemos que $$$u=6 x$$$:

$$\frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} - \frac{5 \cos{\left({\color{red}{u}} \right)}}{96} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(8 x \right)}}{256} - \frac{5 \cos{\left({\color{red}{\left(6 x\right)}} \right)}}{96}$$

Por lo tanto,

$$\int{\sin{\left(5 x \right)} \cos^{5}{\left(x \right)} d x} = \frac{\sin^{2}{\left(5 x \right)}}{160} - \frac{5 \cos{\left(2 x \right)}}{64} - \frac{5 \cos{\left(4 x \right)}}{64} - \frac{5 \cos{\left(6 x \right)}}{96} - \frac{5 \cos{\left(8 x \right)}}{256}$$

Simplificar:

$$\int{\sin{\left(5 x \right)} \cos^{5}{\left(x \right)} d x} = - \frac{- 24 \sin^{2}{\left(5 x \right)} + 300 \cos{\left(2 x \right)} + 300 \cos{\left(4 x \right)} + 200 \cos{\left(6 x \right)} + 75 \cos{\left(8 x \right)}}{3840}$$

Añade la constante de integración:

$$\int{\sin{\left(5 x \right)} \cos^{5}{\left(x \right)} d x} = - \frac{- 24 \sin^{2}{\left(5 x \right)} + 300 \cos{\left(2 x \right)} + 300 \cos{\left(4 x \right)} + 200 \cos{\left(6 x \right)} + 75 \cos{\left(8 x \right)}}{3840}+C$$

$$$\int \sin{\left(5 x \right)} \cos^{5}{\left(x \right)}\, dx = - \frac{- 24 \sin^{2}{\left(5 x \right)} + 300 \cos{\left(2 x \right)} + 300 \cos{\left(4 x \right)} + 200 \cos{\left(6 x \right)} + 75 \cos{\left(8 x \right)}}{3840} + C$$$A