integral from 0 to 2pi of sin(x+1)

Lösung

\int_{0}^{2 π} sin (x + 1) d x

- cos (2 π + 1) + cos (1)

Dezimale

0

Schritte zur Lösung

\int_{0}^{2 π} sin (x + 1) d x

Wende U-Substitution an

= \int_{1}^{2 π + 1} sin (u) d u

Nutze das gemeinsame Integral :

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

= [- cos (u)]_{1}^{2 π + 1}

Berechne die Grenzen:

- cos (2 π + 1) + cos (1)

= - cos (2 π + 1) + cos (1)

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