integral of sec(x)cot^3(x)

Lösung

\int sec (x) cot^{3} (x) d x

\frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

Schritte zur Lösung

\int sec (x) cot^{3} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int \frac{cos ^{2} ( x )}{sin ^{3} ( x )} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

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

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= \frac{1}{4} (\frac{u ^{2}}{2} - 2 ln ∣ u ∣ - \frac{1}{2 u ^{2}})

Setze in

u = tan (\frac{x}{2})

ein

= \frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} - 2 ln tan (\frac{x}{2}) - \frac{1}{2 tan ^{2} ( \frac{x}{2} )})

Vereinfache

\frac{1}{4} (\frac{tan ^{2} ( \frac{x}{2} )}{2} - 2 ln tan (\frac{x}{2}) - \frac{1}{2 tan ^{2} ( \frac{x}{2} )}) : \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

= \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{8} (tan^{2} (\frac{x}{2}) - 4 ln tan (\frac{x}{2}) - cot^{2} (\frac{x}{2})) + C

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