Ολοκλήρωμα του $$$\sin^{3}{\left(x \right)} \cos{\left(2 x \right)}$$$

Βρείτε $$$\int \sin^{3}{\left(x \right)} \cos{\left(2 x \right)}\, dx$$$.

Εφαρμόστε τον τύπο υποβιβασμού δυνάμεων $$$\sin^{3}{\left(\alpha \right)} = \frac{3 \sin{\left(\alpha \right)}}{4} - \frac{\sin{\left(3 \alpha \right)}}{4}$$$ με $$$\alpha=x$$$:

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

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=\frac{1}{4}$$$ και $$$f{\left(x \right)} = \left(3 \sin{\left(x \right)} - \sin{\left(3 x \right)}\right) \cos{\left(2 x \right)}$$$:

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

Expand the expression:

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

Ολοκληρώστε όρο προς όρο:

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

Επαναγράψτε τον ολοκληρωτέο 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 $$$\alpha=3 x$$$ και $$$\beta=2 x$$$:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(3 x \right)} \cos{\left(2 x \right)} d x}}}}{4} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{4}$$

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 $$$f{\left(x \right)} = \sin{\left(x \right)} + \sin{\left(5 x \right)}$$$:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\frac{\sin{\left(x \right)}}{2} + \frac{\sin{\left(5 x \right)}}{2}\right)d x}}}}{4} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}{2}\right)}}}{4}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\left(\sin{\left(x \right)} + \sin{\left(5 x \right)}\right)d x}}}}{8} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\int{\sin{\left(x \right)} d x} + \int{\sin{\left(5 x \right)} d x}\right)}}}{8}$$

Το ολοκλήρωμα του ημιτόνου είναι $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{\int{\sin{\left(5 x \right)} d x}}{8} - \frac{{\color{red}{\int{\sin{\left(x \right)} d x}}}}{8} = \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{\int{\sin{\left(5 x \right)} d x}}{8} - \frac{{\color{red}{\left(- \cos{\left(x \right)}\right)}}}{8}$$

Έστω $$$u=5 x$$$.

Τότε $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{du}{5}$$$.

Επομένως,

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(5 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{5}$$$ και $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{8}$$

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

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{40} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} - \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{40}$$

Θυμηθείτε ότι $$$u=5 x$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{u}} \right)}}{40} = \frac{\cos{\left(x \right)}}{8} + \frac{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}{4} + \frac{\cos{\left({\color{red}{\left(5 x\right)}} \right)}}{40}$$

Επαναγράψτε το $$$\sin\left(x \right)\cos\left(2 x \right)$$$ 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 $$$\alpha=x$$$ και $$$\beta=2 x$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{3 \sin{\left(x \right)} \cos{\left(2 x \right)} d x}}}}{4} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{4}$$

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 $$$f{\left(x \right)} = - 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- \frac{3 \sin{\left(x \right)}}{2} + \frac{3 \sin{\left(3 x \right)}}{2}\right)d x}}}}{4} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(\frac{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}{2}\right)}}}{4}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\left(- 3 \sin{\left(x \right)} + 3 \sin{\left(3 x \right)}\right)d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(- \int{3 \sin{\left(x \right)} d x} + \int{3 \sin{\left(3 x \right)} d x}\right)}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=3$$$ και $$$f{\left(x \right)} = \sin{\left(x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\int{3 \sin{\left(x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{{\color{red}{\left(3 \int{\sin{\left(x \right)} d x}\right)}}}{8}$$

Το ολοκλήρωμα του ημιτόνου είναι $$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\int{\sin{\left(x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40} + \frac{\int{3 \sin{\left(3 x \right)} d x}}{8} - \frac{3 {\color{red}{\left(- \cos{\left(x \right)}\right)}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ με $$$c=3$$$ και $$$f{\left(x \right)} = \sin{\left(3 x \right)}$$$:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{3 \sin{\left(3 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(3 \int{\sin{\left(3 x \right)} d x}\right)}}}{8}$$

Έστω $$$u=3 x$$$.

Τότε $$$du=\left(3 x\right)^{\prime }dx = 3 dx$$$ (τα βήματα παρουσιάζονται »), και έχουμε ότι $$$dx = \frac{du}{3}$$$.

Επομένως,

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\sin{\left(3 x \right)} d x}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{8}$$

Εφαρμόστε τον κανόνα του σταθερού πολλαπλασίου $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ με $$$c=\frac{1}{3}$$$ και $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{3} d u}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{3 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{3}\right)}}}{8}$$

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

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} + \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

Θυμηθείτε ότι $$$u=3 x$$$:

$$\frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} - \frac{\cos{\left({\color{red}{u}} \right)}}{8} = \frac{\cos{\left(x \right)}}{2} + \frac{\cos{\left(5 x \right)}}{40} - \frac{\cos{\left({\color{red}{\left(3 x\right)}} \right)}}{8}$$

Επομένως,

$$\int{\sin^{3}{\left(x \right)} \cos{\left(2 x \right)} d x} = \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\sin^{3}{\left(x \right)} \cos{\left(2 x \right)} d x} = \frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}+C$$

$$$\int \sin^{3}{\left(x \right)} \cos{\left(2 x \right)}\, dx = \left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{8} + \frac{\cos{\left(5 x \right)}}{40}\right) + C$$$A