integral of tan^5(7x)

Lösung

\int tan^{5} (7 x) d x

\frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28} + C

Schritte zur Lösung

\int tan^{5} (7 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (- 1 + sec^{2} (7 x))^{2} tan (7 x) d x

Wende U-Substitution an

= \int \frac{( - 1 + u ^{2} ) ^{2}}{7 u} d u

Multipliziere aus

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

= \int \frac{1}{7 u} - \frac{2 u}{7} + \frac{u ^{3}}{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}{7 u} d u - \int \frac{2 u}{7} d u + \int \frac{u ^{3}}{7} d u

\int \frac{1}{7 u} d u = \frac{1}{7} ln ∣ u ∣

\int \frac{2 u}{7} d u = \frac{u ^{2}}{7}

\int \frac{u ^{3}}{7} d u = \frac{u ^{4}}{28}

= \frac{1}{7} ln ∣ u ∣ - \frac{u ^{2}}{7} + \frac{u ^{4}}{28}

Setze in

u = sec (7 x)

ein

= \frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{7} ln ∣ sec (7 x) ∣ - \frac{sec ^{2} ( 7 x )}{7} + \frac{sec ^{4} ( 7 x )}{28} + C

\frac{d}{d x} (\frac{10}{x - 6})

f (x) = \frac{x}{x ^{2} - x + 1}

x \to 0 + lim (x \cdot sin (\frac{1}{x}))

(3 sin (x))^{^{'}}

a re a f (x) = 2 sin (x), g (x) = 2 cos (x)