integral from 0 to pi+2i of cos(z/2)

Lösung

\int_{0}^{π + 2 i} cos (\frac{z}{2}) d z

2 sin (\frac{2 i + π}{2})

Schritte zur Lösung

\int_{0}^{π + 2 i} cos (\frac{z}{2}) d z

Wende U-Substitution an

= \int_{0}^{\frac{π + 2 i}{2}} cos (u) \cdot 2 d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{\frac{π + 2 i}{2}} cos (u) d u

Nutze das gemeinsame Integral :

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

= 2 [sin (u)]_{0}^{\frac{π + 2 i}{2}}

Berechne die Grenzen:

sin (\frac{2 i + π}{2})

= 2 sin (\frac{2 i + π}{2})

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