integral from 0 to pi/2 of θ^2sin(2θ)

Lösung

\int_{0}^{\frac{π}{2}} θ^{2} sin (2 θ) d θ

\frac{π ^{2} - 2}{8} - \frac{1}{4}

Dezimale

0.73370 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} θ^{2} sin (2 θ) d θ

Wende die partielle Integration an

= [\frac{1}{2} (- θ^{2} cos (2 θ) - 2 \cdot \int - θ cos (2 θ) d θ)]_{0}^{\frac{π}{2}}

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

= [\frac{1}{2} (- θ^{2} cos (2 θ) - 2 (- \frac{1}{4} (2 θ sin (2 θ) + cos (2 θ))))]_{0}^{\frac{π}{2}}

Vereinfache

[\frac{1}{2} (- θ^{2} cos (2 θ) - 2 (- \frac{1}{4} (2 θ sin (2 θ) + cos (2 θ))))]_{0}^{\frac{π}{2}} : [\frac{1}{4} (- 2 θ^{2} cos (2 θ) + 2 θ sin (2 θ) + cos (2 θ))]_{0}^{\frac{π}{2}}

= [\frac{1}{4} (- 2 θ^{2} cos (2 θ) + 2 θ sin (2 θ) + cos (2 θ))]_{0}^{\frac{π}{2}}

Berechne die Grenzen:

\frac{π ^{2} - 2}{8} - \frac{1}{4}

= \frac{π ^{2} - 2}{8} - \frac{1}{4}

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