integral from 0 to pi/4 of tcos(2t)

Lösung

\int_{0}^{\frac{π}{4}} t cos (2 t) d t

\frac{π - 2}{8}

Dezimale

0.14269 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{4}} t cos (2 t) d t

Wende U-Substitution an

= \int_{0}^{\frac{π}{2}} \frac{u cos ( u )}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{\frac{π}{2}} u cos (u) d u

Wende die partielle Integration an

= \frac{1}{4} [u sin (u) - \int sin (u) d u]_{0}^{\frac{π}{2}}

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

= \frac{1}{4} [u sin (u) - (- cos (u))]_{0}^{\frac{π}{2}}

Vereinfache

= \frac{1}{4} [u sin (u) + cos (u)]_{0}^{\frac{π}{2}}

Berechne die Grenzen:

\frac{π}{2} - 1

= \frac{1}{4} (\frac{π}{2} - 1)

Vereinfache

= \frac{π - 2}{8}

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