Ολοκλήρωμα του $$$\sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)}$$$

Βρείτε $$$\int \sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx$$$.

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 $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \sin^{2}{\left(x \right)}\right) \sin^{4}{\left(x \right)} \cos{\left(x \right)} d x}}}$$

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,

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

Expand the expression:

$${\color{red}{\int{u^{4} \left(1 - u^{2}\right) d u}}} = {\color{red}{\int{\left(- u^{6} + u^{4}\right)d u}}}$$

Ολοκληρώστε όρο προς όρο:

$${\color{red}{\int{\left(- u^{6} + u^{4}\right)d u}}} = {\color{red}{\left(\int{u^{4} d u} - \int{u^{6} d u}\right)}}$$

Εφαρμόστε τον κανόνα δύναμης $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=4$$$:

$$- \int{u^{6} d u} + {\color{red}{\int{u^{4} d u}}}=- \int{u^{6} d u} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- \int{u^{6} d u} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Εφαρμόστε τον κανόνα δύναμης $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ με $$$n=6$$$:

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

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

$$\frac{{\color{red}{u}}^{5}}{5} - \frac{{\color{red}{u}}^{7}}{7} = \frac{{\color{red}{\sin{\left(x \right)}}}^{5}}{5} - \frac{{\color{red}{\sin{\left(x \right)}}}^{7}}{7}$$

Επομένως,

$$\int{\sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)} d x} = - \frac{\sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}+C$$

$$$\int \sin^{4}{\left(x \right)} \cos^{3}{\left(x \right)}\, dx = \left(- \frac{\sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}\right) + C$$$A