integral from 0 to 2pi of cos^3(θ)

Lösung

\int_{0}^{2 π} cos^{3} (θ) d θ

0

Schritte zur Lösung

\int_{0}^{2 π} cos^{3} (θ) d θ

Vereinfache

cos^{3} (θ) : cos^{2} (θ) cos (θ)

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

Umschreiben mit Hilfe von Trigonometrie-Identitäten

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

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

Berechne die Grenzen:

0

= 0

\frac{d}{d x} (\frac{3 x ^{2} - 4 x + 5}{6 x})

x \frac{d y}{d x} + y = 9 x y^{2}

x \to 2 + lim (\frac{1}{( x - 2 ) ^{2}})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + 81})

y^{^{'}} - 2 y = 3 x^{2}