integral of cot^{-9}(x)

Lösung

\int cot^{- 9} (x) d x

ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

Schritte zur Lösung

\int cot^{- 9} (x) d x

Vereinfache

cot^{- 9} (x) : tan^{9} (x)

= \int tan^{9} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (- 1 + sec^{2} (x))^{4} tan (x) d x

Wende U-Substitution an

= \int \frac{( - 1 + u ^{2} ) ^{4}}{u} d u

Multipliziere aus

\frac{( - 1 + u ^{2} ) ^{4}}{u} : \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7}

= \int \frac{1}{u} - 4 u + 6 u^{3} - 4 u^{5} + u^{7} 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 \frac{1}{u} d u - \int 4 u d u + \int 6 u^{3} d u - \int 4 u^{5} d u + \int u^{7} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int 4 u d u = 2 u^{2}

\int 6 u^{3} d u = \frac{3 u ^{4}}{2}

\int 4 u^{5} d u = \frac{2 u ^{6}}{3}

\int u^{7} d u = \frac{u ^{8}}{8}

= ln ∣ u ∣ - 2 u^{2} + \frac{3 u ^{4}}{2} - \frac{2 u ^{6}}{3} + \frac{u ^{8}}{8}

Setze in

u = sec (x)

ein

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8}

Füge eine Konstante zur Lösung hinzu

= ln ∣ sec (x) ∣ - 2 sec^{2} (x) + \frac{3 sec ^{4} ( x )}{2} - \frac{2 sec ^{6} ( x )}{3} + \frac{sec ^{8} ( x )}{8} + C

\frac{d}{d x} (1 - 2 x^{2})

\int \frac{x}{( 16 - x ^{2} ) ^{2}} d x

t an g e n t y = \frac{- 4 x}{x ^{2} + 1}, (1, - 2)

\int \frac{10}{x + 10 + x} d x

\int_{2}^{1} (12 x^{- 5}) d x