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

Lösung

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

\frac{2}{3} (\frac{tan ^{3} ( \frac{3 x}{2} )}{3} + \frac{tan ^{5} ( \frac{3 x}{2} )}{5}) + C

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende U-Substitution an

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

Multipliziere aus

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

= \frac{2}{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{2}{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{2}{3} (\frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Ersetze zurück

= \frac{2}{3} (\frac{tan ^{3} ( \frac{3 x}{2} )}{3} + \frac{tan ^{5} ( \frac{3 x}{2} )}{5})

Füge eine Konstante zur Lösung hinzu

= \frac{2}{3} (\frac{tan ^{3} ( \frac{3 x}{2} )}{3} + \frac{tan ^{5} ( \frac{3 x}{2} )}{5}) + C

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