integral of tan^{11}(x)sec^8(x)

Lösung

\int tan^{11} (x) sec^{8} (x) d x

\frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

Schritte zur Lösung

\int tan^{11} (x) sec^{8} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int sec^{2} (x) (1 + tan^{2} (x))^{3} tan^{11} (x) d x

Wende U-Substitution an

= \int u^{11} (1 + u^{2})^{3} d u

Multipliziere aus

u^{11} (1 + u^{2})^{3} : u^{11} + 3 u^{13} + 3 u^{15} + u^{17}

= \int u^{11} + 3 u^{13} + 3 u^{15} + u^{17} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{11} d u + \int 3 u^{13} d u + \int 3 u^{15} d u + \int u^{17} d u

\int u^{11} d u = \frac{u ^{12}}{12}

\int 3 u^{13} d u = \frac{3 u ^{14}}{14}

\int 3 u^{15} d u = \frac{3 u ^{16}}{16}

\int u^{17} d u = \frac{u ^{18}}{18}

= \frac{u ^{12}}{12} + \frac{3 u ^{14}}{14} + \frac{3 u ^{16}}{16} + \frac{u ^{18}}{18}

Setze in

u = tan (x)

ein

= \frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18}

Füge eine Konstante zur Lösung hinzu

= \frac{tan ^{12} ( x )}{12} + \frac{3 tan ^{14} ( x )}{14} + \frac{3 tan ^{16} ( x )}{16} + \frac{tan ^{18} ( x )}{18} + C

d er i v a t i v e (2 + x)^{2}

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