integral from 0 to 2pi of 2cos(2x+4)

Lösung

\int_{0}^{2 π} 2 cos (2 x + 4) d x

sin (4 π + 4) - sin (4)

Dezimale

0

Schritte zur Lösung

\int_{0}^{2 π} 2 cos (2 x + 4) d x

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{2 π} cos (2 x + 4) d x

Wende U-Substitution an

= 2 \cdot \int_{4}^{4 π + 4} cos (u) \frac{1}{2} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{2} \cdot \int_{4}^{4 π + 4} cos (u) d u

Nutze das gemeinsame Integral :

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

= 2 \cdot \frac{1}{2} [sin (u)]_{4}^{4 π + 4}

Vereinfache

2 \cdot \frac{1}{2} [sin (u)]_{4}^{4 π + 4} : [sin (u)]_{4}^{4 π + 4}

= [sin (u)]_{4}^{4 π + 4}

Berechne die Grenzen:

sin (4 π + 4) - sin (4)

= sin (4 π + 4) - sin (4)

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