integral from 0 to pi/2 of sin(nx)

Lösung

\int_{0}^{\frac{π}{2}} sin (n x) d x

\frac{- cos ( \frac{πn}{2} ) + 1}{n}

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} sin (n x) d x

Wende U-Substitution an

= \int_{0}^{\frac{πn}{2}} \frac{sin ( u )}{n} d u

Entferne die Konstante:

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

= \frac{1}{n} \cdot \int_{0}^{\frac{πn}{2}} sin (u) d u

Nutze das gemeinsame Integral :

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

= \frac{1}{n} [- cos (u)]_{0}^{\frac{πn}{2}}

Berechne die Grenzen:

- cos (\frac{πn}{2}) + 1

= \frac{1}{n} (- cos (\frac{πn}{2}) + 1)

Vereinfache

= \frac{- cos ( \frac{πn}{2} ) + 1}{n}

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