integral from 0 to t of ue^{-4u}cos(5u)

Lösung

\int_{0}^{t} u e^{- 4 u} cos (5 u) d u

\frac{5 t e ^{- 4 t} sin ( 5 t ) - 4 t e ^{- 4 t} cos ( 5 t )}{41} + \frac{9}{1681} e^{- 4 t} cos (5 t) + \frac{40}{1681} e^{- 4 t} sin (5 t) - \frac{9}{1681}

Schritte zur Lösung

\int_{0}^{t} u e^{- 4 u} cos (5 u) d u

Wende die partielle Integration an

= [u (\frac{5}{41} e^{- 4 u} sin (5 u) - \frac{4}{41} e^{- 4 u} cos (5 u)) - \int \frac{1}{41} (5 e^{- 4 u} sin (5 u) - 4 e^{- 4 u} cos (5 u)) d u]_{0}^{t}

\int \frac{1}{41} (5 e^{- 4 u} sin (5 u) - 4 e^{- 4 u} cos (5 u)) d u = \frac{1}{1681} (- 9 e^{- 4 u} cos (5 u) - 40 e^{- 4 u} sin (5 u))

= [u (\frac{5}{41} e^{- 4 u} sin (5 u) - \frac{4}{41} e^{- 4 u} cos (5 u)) - \frac{1}{1681} (- 9 e^{- 4 u} cos (5 u) - 40 e^{- 4 u} sin (5 u))]_{0}^{t}

Berechne die Grenzen:

t (\frac{5}{41} e^{- 4 t} sin (5 t) - \frac{4}{41} e^{- 4 t} cos (5 t)) - \frac{1}{1681} (- 9 e^{- 4 t} cos (5 t) - 40 e^{- 4 t} sin (5 t)) - \frac{9}{1681}

= t (\frac{5}{41} e^{- 4 t} sin (5 t) - \frac{4}{41} e^{- 4 t} cos (5 t)) - \frac{1}{1681} (- 9 e^{- 4 t} cos (5 t) - 40 e^{- 4 t} sin (5 t)) - \frac{9}{1681}

Vereinfache

= \frac{5 t e ^{- 4 t} sin ( 5 t ) - 4 t e ^{- 4 t} cos ( 5 t )}{41} + \frac{9}{1681} e^{- 4 t} cos (5 t) + \frac{40}{1681} e^{- 4 t} sin (5 t) - \frac{9}{1681}

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