integral from 0 to pi/2 of xcos(9x)

Lösung

\int_{0}^{\frac{π}{2}} x cos (9 x) d x

\frac{9 π - 2}{162}

Dezimale

0.16218 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{2}} x cos (9 x) d x

Wende U-Substitution an

= \int_{0}^{\frac{9 π}{2}} \frac{u cos ( u )}{81} d u

Entferne die Konstante:

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

= \frac{1}{81} \cdot \int_{0}^{\frac{9 π}{2}} u cos (u) d u

Wende die partielle Integration an

= \frac{1}{81} [u sin (u) - \int sin (u) d u]_{0}^{\frac{9 π}{2}}

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

= \frac{1}{81} [u sin (u) - (- cos (u))]_{0}^{\frac{9 π}{2}}

Vereinfache

= \frac{1}{81} [u sin (u) + cos (u)]_{0}^{\frac{9 π}{2}}

Berechne die Grenzen:

\frac{9 π}{2} - 1

= \frac{1}{81} (\frac{9 π}{2} - 1)

Vereinfache

= \frac{9 π - 2}{162}

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