integral from 0 to pi/4 of sin^5(x)

Lösung

\int_{0}^{\frac{π}{4}} sin^{5} (x) d x

\frac{8}{15} - \frac{43}{60 2}

Dezimale

0.02657 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} sin^{5} (x) d x

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{\frac{π}{4}} (1 - cos^{2} (x))^{2} sin (x) d x

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

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

\int_{\frac{2}{2}}^{1} 2 u^{2} d u = 2 (\frac{1}{3} - \frac{1}{6 2})

\int_{\frac{2}{2}}^{1} u^{4} d u = \frac{1}{5} - \frac{1}{20 2}

= - (- (1 - \frac{1}{2}) + 2 (\frac{1}{3} - \frac{1}{6 2}) - (\frac{1}{5} - \frac{1}{20 2}))

Vereinfache

- (- (1 - \frac{1}{2}) + 2 (\frac{1}{3} - \frac{1}{6 2}) - (\frac{1}{5} - \frac{1}{20 2})) : \frac{8}{15} - \frac{43}{60 2}

= \frac{8}{15} - \frac{43}{60 2}

\int_{- 4}^{0} x^{2} + 4 x d x

\int_{- π}^{0} - 2 cos^{2} (x) sin (x) d x

\int_{0}^{10} 25 - (t - 5)^{2} d t

\int_{2}^{5} (x^{3} - x^{2} + 8) d x

\int_{0}^{4} - x^{2} - 3 x d x