integral from 0 to 2pi of cos^3(t)

Lösung

\int_{0}^{2 π} cos^{3} (t) d t

0

Schritte zur Lösung

\int_{0}^{2 π} cos^{3} (t) d t

Vereinfache

cos^{3} (t) : cos^{2} (t) cos (t)

= \int_{0}^{2 π} cos^{2} (t) cos (t) d t

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende die partielle Integration an

= [- \int - sin (2 t) sin (t) d t + cos^{2} (t) sin (t)]_{0}^{2 π}

\int - sin (2 t) sin (t) d t = - \frac{2}{3} sin^{3} (t)

= [- (- \frac{2}{3} sin^{3} (t)) + cos^{2} (t) sin (t)]_{0}^{2 π}

Vereinfache

[- (- \frac{2}{3} sin^{3} (t)) + cos^{2} (t) sin (t)]_{0}^{2 π} : [\frac{1}{3} (2 sin^{3} (t) + 3 cos^{2} (t) sin (t))]_{0}^{2 π}

= [\frac{1}{3} (2 sin^{3} (t) + 3 cos^{2} (t) sin (t))]_{0}^{2 π}

Berechne die Grenzen:

0

= 0

\int_{0}^{\frac{π}{2}} cos^{11} (x) d x

\int_{0}^{3} x^{6} d x

\int_{0}^{1} e^{2 x^{2}} d x

\int_{4}^{9} x (2 x^{2} + 1)^{6} d x

\int_{1}^{2} \frac{1}{sin ( x ) cos ( x )} d x