integral from 0 to pi of sin(x+kx)

Lösung

\int_{0}^{π} sin (x + k x) d x

\frac{- cos ( π + πk ) + 1}{1 + k}

Schritte zur Lösung

\int_{0}^{π} sin (x + k x) d x

Wende U-Substitution an

= \int_{0}^{π + kπ} \frac{sin ( u )}{1 + k} d u

Entferne die Konstante:

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

= \frac{1}{1 + k} \cdot \int_{0}^{π + kπ} sin (u) d u

Nutze das gemeinsame Integral :

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

= \frac{1}{1 + k} [- cos (u)]_{0}^{π + kπ}

Berechne die Grenzen:

- cos (π + πk) + 1

= \frac{1}{1 + k} (- cos (π + πk) + 1)

Vereinfache

= \frac{- cos ( π + πk ) + 1}{1 + k}

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