Ολοκλήρωμα του $$$\left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}$$$

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

Απλοποιήστε τον ολοκληρωτέο:

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

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 $$$f{\left(x \right)} = \left(- \sin{\left(x \right)} + \cos{\left(x \right)}\right)^{2}$$$:

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

Expand the expression:

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

Ολοκληρώστε όρο προς όρο:

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

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

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

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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$:

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

Ολοκληρώστε όρο προς όρο:

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

Εφαρμόστε τον κανόνα της σταθεράς $$$\int c\, dx = c x$$$ με $$$c=1$$$:

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

Έστω $$$u=2 x$$$.

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

Επομένως,

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

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

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

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

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

Θυμηθείτε ότι $$$u=2 x$$$:

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

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

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\sin^{2}{\left(x \right)} d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4}$$

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 $$$f{\left(x \right)} = 1 - \cos{\left(2 x \right)}$$$:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}\right)d x}}}}{4} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}{2}\right)}}}{4}$$

Ολοκληρώστε όρο προς όρο:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\int{\left(1 - \cos{\left(2 x \right)}\right)d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} + \frac{{\color{red}{\left(\int{1 d x} - \int{\cos{\left(2 x \right)} d x}\right)}}}{8}$$

Εφαρμόστε τον κανόνα της σταθεράς $$$\int c\, dx = c x$$$ με $$$c=1$$$:

$$\frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{\int{1 d x}}}}{8} = \frac{x}{8} + \frac{\sin{\left(2 x \right)}}{16} - \frac{\int{2 \sin{\left(x \right)} \cos{\left(x \right)} d x}}{4} - \frac{\int{\cos{\left(2 x \right)} d x}}{8} + \frac{{\color{red}{x}}}{8}$$

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

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

Επομένως,

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

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

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

Έστω $$$v=\sin{\left(x \right)}$$$.

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

Το ολοκλήρωμα γίνεται

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

Εφαρμόστε τον κανόνα δύναμης $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=1$$$:

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

Θυμηθείτε ότι $$$v=\sin{\left(x \right)}$$$:

$$\frac{x}{4} - \frac{{\color{red}{v}}^{2}}{4} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(x \right)}}}^{2}}{4}$$

Επομένως,

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

Απλοποιήστε:

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

Προσθέστε τη σταθερά ολοκλήρωσης:

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

$$$\int \left(- \frac{\sin{\left(x \right)}}{2} + \frac{\cos{\left(x \right)}}{2}\right)^{2}\, dx = \frac{x - \sin^{2}{\left(x \right)}}{4} + C$$$A