integral from 0 to 2pi of xcos(2x)

Lösung

\int_{0}^{2 π} x cos (2 x) d x

0

Schritte zur Lösung

\int_{0}^{2 π} x cos (2 x) d x

Wende U-Substitution an

= \int_{0}^{4 π} \frac{u cos ( u )}{4} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{0}^{4 π} u cos (u) d u

Wende die partielle Integration an

= \frac{1}{4} [u sin (u) - \int sin (u) d u]_{0}^{4 π}

\int sin (u) d u = - cos (u)

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

Vereinfache

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

Berechne die Grenzen:

0

= \frac{1}{4} \cdot 0

Vereinfache

= 0

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