integral from 0 to pi of cos^4(θ)

Lösung

\int_{0}^{π} cos^{4} (θ) d θ

\frac{3 π}{8}

Dezimale

1.17809 \dots

Schritte zur Lösung

\int_{0}^{π} cos^{4} (θ) d θ

Vereinfache

cos^{4} (θ) : cos^{3} (θ) cos (θ)

= \int_{0}^{π} cos^{3} (θ) cos (θ) d θ

Wende die partielle Integration an

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

\int - 3 cos^{2} (θ) sin^{2} (θ) d θ = - \frac{3}{8} (θ - \frac{1}{4} sin (4 θ))

= [cos^{3} (θ) sin (θ) - (- \frac{3}{8} (θ - \frac{1}{4} sin (4 θ)))]_{0}^{π}

Vereinfache

[cos^{3} (θ) sin (θ) - (- \frac{3}{8} (θ - \frac{1}{4} sin (4 θ)))]_{0}^{π} : [\frac{1}{32} (32 cos^{3} (θ) sin (θ) + 3 (4 θ - sin (4 θ)))]_{0}^{π}

= [\frac{1}{32} (32 cos^{3} (θ) sin (θ) + 3 (4 θ - sin (4 θ)))]_{0}^{π}

Berechne die Grenzen:

\frac{3 π}{8}

= \frac{3 π}{8}

\int_{- 1}^{1} (x^{3} - x^{4}) d x

\int_{0}^{1} 3 x^{2} arcsin (x) d x

\int_{- 5}^{2} \frac{5}{x ^{4}} d x

\int_{0}^{2} - \frac{3}{2} x + 3 d x

\int_{1}^{3} (1 + x) d x