Ολοκλήρωμα του $$$\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}$$$

Βρείτε $$$\int \sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.

Απομονώστε ένα συνημίτονο και εκφράστε τα υπόλοιπα σε όρους του ημιτόνου, χρησιμοποιώντας τον τύπο $$$\cos^2\left(\alpha \right)=-\sin^2\left(\alpha \right)+1$$$ με $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{5}{\left(x \right)} \cos{\left(x \right)} d x}}}$$

Έστω $$$u=\sin{\left(x \right)}$$$.

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

Επομένως,

$${\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{5}{\left(x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{u^{5} \left(1 - u^{2}\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{u^{5} \left(1 - u^{2}\right) d u}}} = {\color{red}{\int{\left(- u^{7} + u^{5}\right)d u}}}$$

Ολοκληρώστε όρο προς όρο:

$${\color{red}{\int{\left(- u^{7} + u^{5}\right)d u}}} = {\color{red}{\left(\int{u^{5} d u} - \int{u^{7} d u}\right)}}$$

Εφαρμόστε τον κανόνα δύναμης $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=5$$$:

$$- \int{u^{7} d u} + {\color{red}{\int{u^{5} d u}}}=- \int{u^{7} d u} + {\color{red}{\frac{u^{1 + 5}}{1 + 5}}}=- \int{u^{7} d u} + {\color{red}{\left(\frac{u^{6}}{6}\right)}}$$

Εφαρμόστε τον κανόνα δύναμης $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=7$$$:

$$\frac{u^{6}}{6} - {\color{red}{\int{u^{7} d u}}}=\frac{u^{6}}{6} - {\color{red}{\frac{u^{1 + 7}}{1 + 7}}}=\frac{u^{6}}{6} - {\color{red}{\left(\frac{u^{8}}{8}\right)}}$$

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

$$\frac{{\color{red}{u}}^{6}}{6} - \frac{{\color{red}{u}}^{8}}{8} = \frac{{\color{red}{\sin{\left(x \right)}}}^{6}}{6} - \frac{{\color{red}{\sin{\left(x \right)}}}^{8}}{8}$$

Επομένως,

$$\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}+C$$

$$$\int \sin^{5}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(- \frac{\sin^{8}{\left(x \right)}}{8} + \frac{\sin^{6}{\left(x \right)}}{6}\right) + C$$$A