integral from 0 to pi/3 of tan^5(x)

Lösung

\int_{0}^{\frac{π}{3}} tan^{5} (x) d x

ln (2) + \frac{3}{4}

Dezimale

1.44314 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{3}} tan^{5} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{3}} (- 1 + sec^{2} (x))^{2} tan (x) d x

Wende U-Substitution an

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

Multipliziere aus

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

= \int_{1}^{2} \frac{1}{u} - 2 u + 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

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

\int_{1}^{2} \frac{1}{u} d u = ln (2)

\int_{1}^{2} 2 u d u = 3

\int_{1}^{2} u^{3} d u = \frac{15}{4}

= ln (2) - 3 + \frac{15}{4}

Ziehe Brüche zusammen

- 3 + \frac{15}{4} : \frac{3}{4}

= ln (2) + \frac{3}{4}

y^{^{''}} + \frac{6}{x} y^{^{'}} + \frac{4}{x ^{2}} y = 0

d er i v a t i v e f (x) = x^{2} - 4 x + 2

x \to 1 lim (e^{(x^{3} - x)})

\frac{d}{d x} (e^{π x})

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