Ολοκλήρωμα του $$$\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)}$$$

Βρείτε $$$\int \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)}\, dx$$$.

Επαναγράψτε το $$$\cos\left(2 x \right)\cos\left(6 x \right)$$$ χρησιμοποιώντας τον τύπο $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ με $$$\alpha=2 x$$$ και $$$\beta=6 x$$$:

$${\color{red}{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{\cos{\left(8 x \right)}}{2}\right) \sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}}$$

Αναπτύξτε την παράσταση:

$${\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} + \frac{\cos{\left(8 x \right)}}{2}\right) \sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}} = {\color{red}{\int{\left(\frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)}}{2} + \frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)}}{2}\right)d x}}}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(x \right)} = \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)} + \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)}$$$:

$${\color{red}{\int{\left(\frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)}}{2} + \frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)} + \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)}\right)d x}}{2}\right)}}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{{\color{red}{\int{\left(\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)} + \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)} d x} + \int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}\right)}}}{2}$$

Επαναγράψτε το $$$\sin\left(2 x \right)\cos\left(4 x \right)$$$ χρησιμοποιώντας τον τύπο $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ με $$$\alpha=2 x$$$ και $$$\beta=4 x$$$:

$$\frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(4 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}\right) \sin{\left(6 x \right)} d x}}}}{2}$$

Αναπτύξτε την παράσταση:

$$\frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)}}{2} + \frac{\sin{\left(6 x \right)}}{2}\right) \sin{\left(6 x \right)} d x}}}}{2} = \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)}}{2} + \frac{\sin^{2}{\left(6 x \right)}}{2}\right)d x}}}}{2}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(x \right)} = - \sin{\left(2 x \right)} \sin{\left(6 x \right)} + \sin^{2}{\left(6 x \right)}$$$:

$$\frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(2 x \right)} \sin{\left(6 x \right)}}{2} + \frac{\sin^{2}{\left(6 x \right)}}{2}\right)d x}}}}{2} = \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(- \sin{\left(2 x \right)} \sin{\left(6 x \right)} + \sin^{2}{\left(6 x \right)}\right)d x}}{2}\right)}}}{2}$$

Ολοκληρώστε όρο προς όρο:

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

Έστω $$$u=6 x$$$.

Τότε $$$du=\left(6 x\right)^{\prime }dx = 6 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{du}{6}$$$.

Επομένως,

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

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{6}$$$ και $$$f{\left(u \right)} = \sin^{2}{\left(u \right)}$$$:

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

Εφαρμόστε τον τύπο υποβιβασμού δυνάμεων $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ με $$$\alpha= u $$$:

$$- \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\sin^{2}{\left(u \right)} d u}}}}{24} = - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{24}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(u \right)} = 1 - \cos{\left(2 u \right)}$$$:

$$- \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 u \right)}}{2}\right)d u}}}}{24} = - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}{2}\right)}}}{24}$$

Ολοκληρώστε όρο προς όρο:

$$- \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(1 - \cos{\left(2 u \right)}\right)d u}}}}{48} = - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} + \frac{{\color{red}{\left(\int{1 d u} - \int{\cos{\left(2 u \right)} d u}\right)}}}{48}$$

Εφαρμόστε τον κανόνα της σταθεράς $$$\int c\, du = c u$$$ με $$$c=1$$$:

$$- \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(2 u \right)} d u}}{48} + \frac{{\color{red}{\int{1 d u}}}}{48} = - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(2 u \right)} d u}}{48} + \frac{{\color{red}{u}}}{48}$$

Έστω $$$v=2 u$$$.

Τότε $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$du = \frac{dv}{2}$$$.

Επομένως,

$$\frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(2 u \right)} d u}}}}{48} = \frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{48}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$\frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{2} d v}}}}{48} = \frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{2}\right)}}}{48}$$

Το ολοκλήρωμα του συνημιτόνου είναι $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$\frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{96} = \frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\sin{\left(v \right)}}}}{96}$$

Θυμηθείτε ότι $$$v=2 u$$$:

$$\frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{v}} \right)}}{96} = \frac{u}{48} - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(2 u\right)}} \right)}}{96}$$

Θυμηθείτε ότι $$$u=6 x$$$:

$$- \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left(2 {\color{red}{u}} \right)}}{96} + \frac{{\color{red}{u}}}{48} = - \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}{4} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left(2 {\color{red}{\left(6 x\right)}} \right)}}{96} + \frac{{\color{red}{\left(6 x\right)}}}{48}$$

Ξαναγράψτε τον ολοκληρωτέο χρησιμοποιώντας τον τύπο $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ με $$$\alpha=2 x$$$ και $$$\beta=6 x$$$:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} d x}}}}{4} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} - \frac{\cos{\left(8 x \right)}}{2}\right)d x}}}}{4}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(x \right)} = \cos{\left(4 x \right)} - \cos{\left(8 x \right)}$$$:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} - \frac{\cos{\left(8 x \right)}}{2}\right)d x}}}}{4} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(4 x \right)} - \cos{\left(8 x \right)}\right)d x}}{2}\right)}}}{4}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\left(\cos{\left(4 x \right)} - \cos{\left(8 x \right)}\right)d x}}}}{8} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\left(\int{\cos{\left(4 x \right)} d x} - \int{\cos{\left(8 x \right)} d x}\right)}}}{8}$$

Έστω $$$u=8 x$$$.

Τότε $$$du=\left(8 x\right)^{\prime }dx = 8 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{du}{8}$$$.

Το ολοκλήρωμα μπορεί να επαναγραφεί ως

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(8 x \right)} d x}}}}{8} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{8}$$$ και $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{8} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{8}$$

Το ολοκλήρωμα του συνημιτόνου είναι $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{64} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{{\color{red}{\sin{\left(u \right)}}}}{64}$$

Θυμηθείτε ότι $$$u=8 x$$$:

$$\frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{u}} \right)}}{64} = \frac{x}{8} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\int{\cos{\left(4 x \right)} d x}}{8} + \frac{\sin{\left({\color{red}{\left(8 x\right)}} \right)}}{64}$$

Έστω $$$u=4 x$$$.

Τότε $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{du}{4}$$$.

Επομένως,

$$\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{4}$$$ και $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{8} = \frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{8}$$

Το ολοκλήρωμα του συνημιτόνου είναι $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{32} = \frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{32}$$

Θυμηθείτε ότι $$$u=4 x$$$:

$$\frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{32} = \frac{x}{8} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}{2} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{32}$$

Επαναγράψτε το $$$\sin\left(2 x \right)\cos\left(8 x \right)$$$ χρησιμοποιώντας τον τύπο $$$\sin\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \sin\left(\alpha-\beta \right)+\frac{1}{2} \sin\left(\alpha+\beta \right)$$$ με $$$\alpha=2 x$$$ και $$$\beta=8 x$$$:

$$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(8 x \right)} d x}}}}{2} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(6 x \right)}}{2} + \frac{\sin{\left(10 x \right)}}{2}\right) \sin{\left(6 x \right)} d x}}}}{2}$$

Αναπτύξτε την παράσταση:

$$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\left(- \frac{\sin{\left(6 x \right)}}{2} + \frac{\sin{\left(10 x \right)}}{2}\right) \sin{\left(6 x \right)} d x}}}}{2} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\left(- \frac{\sin^{2}{\left(6 x \right)}}{2} + \frac{\sin{\left(6 x \right)} \sin{\left(10 x \right)}}{2}\right)d x}}}}{2}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(x \right)} = - \sin^{2}{\left(6 x \right)} + \sin{\left(6 x \right)} \sin{\left(10 x \right)}$$$:

$$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\left(- \frac{\sin^{2}{\left(6 x \right)}}{2} + \frac{\sin{\left(6 x \right)} \sin{\left(10 x \right)}}{2}\right)d x}}}}{2} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\left(\frac{\int{\left(- \sin^{2}{\left(6 x \right)} + \sin{\left(6 x \right)} \sin{\left(10 x \right)}\right)d x}}{2}\right)}}}{2}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\int{\left(- \sin^{2}{\left(6 x \right)} + \sin{\left(6 x \right)} \sin{\left(10 x \right)}\right)d x}}}}{4} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{{\color{red}{\left(\int{\sin{\left(6 x \right)} \sin{\left(10 x \right)} d x} - \int{\sin^{2}{\left(6 x \right)} d x}\right)}}}{4}$$

Το ολοκλήρωμα $$$\int{\sin^{2}{\left(6 x \right)} d x}$$$ έχει ήδη υπολογιστεί:

$$\int{\sin^{2}{\left(6 x \right)} d x} = \frac{x}{2} - \frac{\sin{\left(12 x \right)}}{24}$$

Επομένως,

$$\frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(6 x \right)} \sin{\left(10 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin^{2}{\left(6 x \right)} d x}}}}{4} = \frac{x}{8} - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(12 x \right)}}{96} + \frac{\int{\sin{\left(6 x \right)} \sin{\left(10 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{x}{2} - \frac{\sin{\left(12 x \right)}}{24}\right)}}}{4}$$

Ξαναγράψτε τον ολοκληρωτέο χρησιμοποιώντας τον τύπο $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ με $$$\alpha=6 x$$$ και $$$\beta=10 x$$$:

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

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{2}$$$ και $$$f{\left(x \right)} = \cos{\left(4 x \right)} - \cos{\left(16 x \right)}$$$:

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(4 x \right)}}{2} - \frac{\cos{\left(16 x \right)}}{2}\right)d x}}}}{4} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(4 x \right)} - \cos{\left(16 x \right)}\right)d x}}{2}\right)}}}{4}$$

Ολοκληρώστε όρο προς όρο:

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{{\color{red}{\int{\left(\cos{\left(4 x \right)} - \cos{\left(16 x \right)}\right)d x}}}}{8} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{{\color{red}{\left(\int{\cos{\left(4 x \right)} d x} - \int{\cos{\left(16 x \right)} d x}\right)}}}{8}$$

Έστω $$$v=16 x$$$.

Τότε $$$dv=\left(16 x\right)^{\prime }dx = 16 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{dv}{16}$$$.

Το ολοκλήρωμα γίνεται

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(16 x \right)} d x}}}}{8} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{16} d v}}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ με $$$c=\frac{1}{16}$$$ και $$$f{\left(v \right)} = \cos{\left(v \right)}$$$:

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{16} d v}}}}{8} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{16}\right)}}}{8}$$

Το ολοκλήρωμα του συνημιτόνου είναι $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$:

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{128} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{{\color{red}{\sin{\left(v \right)}}}}{128}$$

Θυμηθείτε ότι $$$v=16 x$$$:

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{v}} \right)}}{128} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} + \frac{\int{\cos{\left(4 x \right)} d x}}{8} - \frac{\sin{\left({\color{red}{\left(16 x\right)}} \right)}}{128}$$

Το ολοκλήρωμα $$$\int{\cos{\left(4 x \right)} d x}$$$ έχει ήδη υπολογιστεί:

$$\int{\cos{\left(4 x \right)} d x} = \frac{\sin{\left(4 x \right)}}{4}$$

Επομένως,

$$- \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(16 x \right)}}{128} + \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{8} = - \frac{\sin{\left(4 x \right)}}{32} + \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(16 x \right)}}{128} + \frac{{\color{red}{\left(\frac{\sin{\left(4 x \right)}}{4}\right)}}}{8}$$

Επομένως,

$$\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)} d x} = \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(16 x \right)}}{128}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)} d x} = \frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(16 x \right)}}{128}+C$$

$$$\int \sin{\left(2 x \right)} \sin{\left(6 x \right)} \cos{\left(2 x \right)} \cos{\left(6 x \right)}\, dx = \left(\frac{\sin{\left(8 x \right)}}{64} - \frac{\sin{\left(16 x \right)}}{128}\right) + C$$$A