integral of tan^2(3x)sec^4(3x)

Lösung

\int tan^{2} (3 x) sec^{4} (3 x) d x

\frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

Schritte zur Lösung

\int tan^{2} (3 x) sec^{4} (3 x) d x

Wende U-Substitution an

: \int tan^{2} (u) sec^{4} (u) \frac{1}{3} d u

= \int tan^{2} (u) sec^{4} (u) \frac{1}{3} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int tan^{2} (u) sec^{4} (u) d u

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) tan^{2} (u) d u

Wende U-Substitution an

: \int v^{2} (1 + v^{2}) d v

= \frac{1}{3} \cdot \int v^{2} (1 + v^{2}) d v

Multipliziere aus

v^{2} (1 + v^{2}) : v^{2} + v^{4}

= \frac{1}{3} \cdot \int v^{2} + v^{4} d v

Wende die Summenregel an:

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

= \frac{1}{3} (\int v^{2} d v + \int v^{4} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{3} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Ersetze zurück

= \frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5})

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} (\frac{tan ^{3} ( 3 x )}{3} + \frac{tan ^{5} ( 3 x )}{5}) + C

\int (csc^{2} (x) - 1) cot (x) d x

\frac{d}{d x} (\frac{1}{6 x - 7})

x \to - 1 lim (3 x^{2} - 3 x + 3)

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