Ολοκλήρωμα του $$$\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}$$$

Βρείτε $$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta$$$.

Επαναγράψτε την ολοκληρωτέα χρησιμοποιώντας τον τύπο διπλής γωνίας $$$\sin\left(\theta \right)\cos\left(\theta \right)=\frac{1}{2}\sin\left( 2 \theta \right)$$$:

$${\color{red}{\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta}}} = {\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ με $$$c=\frac{1}{64}$$$ και $$$f{\left(\theta \right)} = \sin^{6}{\left(2 \theta \right)}$$$:

$${\color{red}{\int{\frac{\sin^{6}{\left(2 \theta \right)}}{64} d \theta}}} = {\color{red}{\left(\frac{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}{64}\right)}}$$

Εφαρμόστε τον τύπο υποβιβασμού δυνάμεων $$$\sin^{6}{\left(\alpha \right)} = - \frac{15 \cos{\left(2 \alpha \right)}}{32} + \frac{3 \cos{\left(4 \alpha \right)}}{16} - \frac{\cos{\left(6 \alpha \right)}}{32} + \frac{5}{16}$$$ με $$$\alpha=2 \theta$$$:

$$\frac{{\color{red}{\int{\sin^{6}{\left(2 \theta \right)} d \theta}}}}{64} = \frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ με $$$c=\frac{1}{32}$$$ και $$$f{\left(\theta \right)} = - 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10$$$:

$$\frac{{\color{red}{\int{\left(- \frac{15 \cos{\left(4 \theta \right)}}{32} + \frac{3 \cos{\left(8 \theta \right)}}{16} - \frac{\cos{\left(12 \theta \right)}}{32} + \frac{5}{16}\right)d \theta}}}}{64} = \frac{{\color{red}{\left(\frac{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}{32}\right)}}}{64}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{{\color{red}{\int{\left(- 15 \cos{\left(4 \theta \right)} + 6 \cos{\left(8 \theta \right)} - \cos{\left(12 \theta \right)} + 10\right)d \theta}}}}{2048} = \frac{{\color{red}{\left(\int{10 d \theta} - \int{15 \cos{\left(4 \theta \right)} d \theta} + \int{6 \cos{\left(8 \theta \right)} d \theta} - \int{\cos{\left(12 \theta \right)} d \theta}\right)}}}{2048}$$

Εφαρμόστε τον κανόνα της σταθεράς $$$\int c\, d\theta = c \theta$$$ με $$$c=10$$$:

$$- \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\int{10 d \theta}}}}{2048} = - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\int{\cos{\left(12 \theta \right)} d \theta}}{2048} + \frac{{\color{red}{\left(10 \theta\right)}}}{2048}$$

Έστω $$$u=12 \theta$$$.

Τότε $$$du=\left(12 \theta\right)^{\prime }d\theta = 12 d\theta$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$d\theta = \frac{du}{12}$$$.

Το ολοκλήρωμα μπορεί να επαναγραφεί ως

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(12 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{12}$$$ και $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{12} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{12}\right)}}}{2048}$$

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

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\sin{\left(u \right)}}}}{24576}$$

Θυμηθείτε ότι $$$u=12 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{u}} \right)}}{24576} = \frac{5 \theta}{1024} - \frac{\int{15 \cos{\left(4 \theta \right)} d \theta}}{2048} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{\sin{\left({\color{red}{\left(12 \theta\right)}} \right)}}{24576}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ με $$$c=15$$$ και $$$f{\left(\theta \right)} = \cos{\left(4 \theta \right)}$$$:

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\int{15 \cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{{\color{red}{\left(15 \int{\cos{\left(4 \theta \right)} d \theta}\right)}}}{2048}$$

Έστω $$$u=4 \theta$$$.

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

Επομένως,

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(4 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048}$$

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

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\frac{\cos{\left(u \right)}}{4} d u}}}}{2048} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{4}\right)}}}{2048}$$

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

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 {\color{red}{\sin{\left(u \right)}}}}{8192}$$

Θυμηθείτε ότι $$$u=4 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{\int{6 \cos{\left(8 \theta \right)} d \theta}}{2048} - \frac{15 \sin{\left({\color{red}{\left(4 \theta\right)}} \right)}}{8192}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ με $$$c=6$$$ και $$$f{\left(\theta \right)} = \cos{\left(8 \theta \right)}$$$:

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\int{6 \cos{\left(8 \theta \right)} d \theta}}}}{2048} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{{\color{red}{\left(6 \int{\cos{\left(8 \theta \right)} d \theta}\right)}}}{2048}$$

Έστω $$$u=8 \theta$$$.

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

Επομένως,

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(8 \theta \right)} d \theta}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024}$$

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

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\frac{\cos{\left(u \right)}}{8} d u}}}}{1024} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{8}\right)}}}{1024}$$

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

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\int{\cos{\left(u \right)} d u}}}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 {\color{red}{\sin{\left(u \right)}}}}{8192}$$

Θυμηθείτε ότι $$$u=8 \theta$$$:

$$\frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{u}} \right)}}{8192} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576} + \frac{3 \sin{\left({\color{red}{\left(8 \theta\right)}} \right)}}{8192}$$

Επομένως,

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = \frac{5 \theta}{1024} - \frac{15 \sin{\left(4 \theta \right)}}{8192} + \frac{3 \sin{\left(8 \theta \right)}}{8192} - \frac{\sin{\left(12 \theta \right)}}{24576}$$

Απλοποιήστε:

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)} d \theta} = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576}+C$$

$$$\int \sin^{6}{\left(\theta \right)} \cos^{6}{\left(\theta \right)}\, d\theta = - \frac{- 120 \theta + 45 \sin{\left(4 \theta \right)} - 9 \sin{\left(8 \theta \right)} + \sin{\left(12 \theta \right)}}{24576} + C$$$A