integral of tan^5(4x)

Lösung

\int tan^{5} (4 x) d x

\frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16} + C

Schritte zur Lösung

\int tan^{5} (4 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

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

= \int \frac{1}{4 u} - \frac{u}{2} + \frac{u ^{3}}{4} 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}{4 u} d u - \int \frac{u}{2} d u + \int \frac{u ^{3}}{4} d u

\int \frac{1}{4 u} d u = \frac{1}{4} ln ∣ u ∣

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

\int \frac{u ^{3}}{4} d u = \frac{u ^{4}}{16}

= \frac{1}{4} ln ∣ u ∣ - \frac{u ^{2}}{4} + \frac{u ^{4}}{16}

Setze in

u = sec (4 x)

ein

= \frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln ∣ sec (4 x) ∣ - \frac{sec ^{2} ( 4 x )}{4} + \frac{sec ^{4} ( 4 x )}{16} + C

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

x^{^{''}} + 3 x^{^{'}} + 2 x = 0

d er i v a t i v e x + 4

\frac{d y}{d x} + \frac{6 y}{x} = x

x \to 4 lim (arctan (\frac{x ^{2} - 16}{7 x ^{2} - 28 x}))