integral of cot^{-3}(3y)csc^4(3y)

Lösung

\int cot^{- 3} (3 y) csc^{4} (3 y) d y

\frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y)) + C

Schritte zur Lösung

\int cot^{- 3} (3 y) csc^{4} (3 y) d y

Wende U-Substitution an

= \int \frac{u ^{\frac{2}{3}} + 1}{9 u} d u

Entferne die Konstante:

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

= \frac{1}{9} \cdot \int \frac{u ^{\frac{2}{3}} + 1}{u} d u

Multipliziere aus

\frac{u ^{\frac{2}{3}} + 1}{u} : \frac{1}{u ^{\frac{1}{3}}} + \frac{1}{u}

= \frac{1}{9} \cdot \int \frac{1}{u ^{\frac{1}{3}}} + \frac{1}{u} 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}{9} (\int \frac{1}{u ^{\frac{1}{3}}} d u + \int \frac{1}{u} d u)

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

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{9} (\frac{3}{2} u^{\frac{2}{3}} + ln ∣ u ∣)

Setze in

u = cot^{- 3} (3 y)

ein

= \frac{1}{9} (\frac{3}{2} (cot^{- 3} (3 y))^{\frac{2}{3}} + ln cot^{- 3} (3 y))

Vereinfache

\frac{1}{9} (\frac{3}{2} (cot^{- 3} (3 y))^{\frac{2}{3}} + ln cot^{- 3} (3 y)) : \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y))

= \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{9} (\frac{3}{2} tan^{2} (3 y) + ln tan^{3} (3 y)) + C

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