integral of sec(4x)tan^5(4x)

Lösung

\int sec (4 x) tan^{5} (4 x) d x

\frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20} + C

Schritte zur Lösung

\int sec (4 x) tan^{5} (4 x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int (- 1 + sec^{2} (4 x))^{2} tan (4 x) sec (4 x) d x

Wende U-Substitution an

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

Multipliziere aus

\frac{( - 1 + u ^{2} ) ^{2}}{4} : \frac{1}{4} - \frac{u ^{2}}{2} + \frac{u ^{4}}{4}

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

\int \frac{1}{4} d u = \frac{1}{4} u

\int \frac{u ^{2}}{2} d u = \frac{u ^{3}}{6}

\int \frac{u ^{4}}{4} d u = \frac{u ^{5}}{20}

= \frac{1}{4} u - \frac{u ^{3}}{6} + \frac{u ^{5}}{20}

Setze in

u = sec (4 x)

ein

= \frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20}

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} sec (4 x) - \frac{sec ^{3} ( 4 x )}{6} + \frac{sec ^{5} ( 4 x )}{20} + C

(1 + x^{2}) \frac{d y}{d x} = (1 + y)^{2}

\int (x^{3} + x^{2})^{12} (3 x^{2} + 2 x) d x

y^{^{'}} + x^{2} y = x^{2} y^{4}

\int \frac{5 x ^{2} - 4 x + 4}{x ^{3} + 4 x} d x

\int_{2}^{7} t^{2} ln (3 t) d t