integral of sec^8(2θ)tan^5(2θ)

Lösung

\int sec^{8} (2 θ) tan^{5} (2 θ) d θ

\frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16} + C

Schritte zur Lösung

\int sec^{8} (2 θ) tan^{5} (2 θ) d θ

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (- 1 + sec^{2} (2 θ))^{2} tan (2 θ) sec^{8} (2 θ) d θ

Wende U-Substitution an

= \int \frac{u ^{7} ( u ^{2} - 1 ) ^{2}}{2} d u

Multipliziere aus

\frac{u ^{7} ( u ^{2} - 1 ) ^{2}}{2} : \frac{u ^{11}}{2} - u^{9} + \frac{u ^{7}}{2}

= \int \frac{u ^{11}}{2} - u^{9} + \frac{u ^{7}}{2} 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{u ^{11}}{2} d u - \int u^{9} d u + \int \frac{u ^{7}}{2} d u

\int \frac{u ^{11}}{2} d u = \frac{u ^{12}}{24}

\int u^{9} d u = \frac{u ^{10}}{10}

\int \frac{u ^{7}}{2} d u = \frac{u ^{8}}{16}

= \frac{u ^{12}}{24} - \frac{u ^{10}}{10} + \frac{u ^{8}}{16}

Setze in

u = sec (2 θ)

ein

= \frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16}

Füge eine Konstante zur Lösung hinzu

= \frac{sec ^{12} ( 2 θ )}{24} - \frac{sec ^{10} ( 2 θ )}{10} + \frac{sec ^{8} ( 2 θ )}{16} + C

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