Ολοκλήρωμα του $$$\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}$$$

Βρείτε $$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx$$$.

Πολλαπλασιάστε τον αριθμητή και τον παρονομαστή επί $$$\cos^{4}{\left(x \right)}$$$ και μετατρέψτε το $$$\frac{\sin^{4}{\left(x \right)}}{\cos^{4}{\left(x \right)}}$$$ σε $$$\tan^{4}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{\tan^{4}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}}$$

Μετατρέψτε $$$\frac{1}{\cos^{2}{\left(x \right)}}$$$ σε $$$\sec^{2}{\left(x \right)}$$$:

$${\color{red}{\int{\frac{\tan^{4}{\left(x \right)}}{\cos^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}}$$

Έστω $$$u=\tan{\left(x \right)}$$$.

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

Επομένως,

$${\color{red}{\int{\tan^{4}{\left(x \right)} \sec^{2}{\left(x \right)} d x}}} = {\color{red}{\int{u^{4} d u}}}$$

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

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

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

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

Επομένως,

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}$$

Προσθέστε τη σταθερά ολοκλήρωσης:

$$\int{\frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}} d x} = \frac{\tan^{5}{\left(x \right)}}{5}+C$$

$$$\int \frac{\sin^{4}{\left(x \right)}}{\cos^{6}{\left(x \right)}}\, dx = \frac{\tan^{5}{\left(x \right)}}{5} + C$$$A