integral from 0 to t of (t-u)^2cos(2u)

Lösung

\int_{0}^{t} (t - u)^{2} cos (2 u) d u

\frac{t ^{2} sin ( 2 t )}{2} - \frac{t ( 2 t sin ( 2 t ) + cos ( 2 t ) - 1 )}{2} + \frac{t ^{2} sin ( 2 t ) + t cos ( 2 t )}{2} - \frac{1}{4} sin (2 t)

Schritte zur Lösung

\int_{0}^{t} (t - u)^{2} cos (2 u) d u

Multipliziere aus

(t - u)^{2} cos (2 u) : t^{2} cos (2 u) - 2 t u cos (2 u) + u^{2} cos (2 u)

= \int_{0}^{t} t^{2} cos (2 u) - 2 t u cos (2 u) + u^{2} cos (2 u) d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{t} t^{2} cos (2 u) d u - \int_{0}^{t} 2 t u cos (2 u) d u + \int_{0}^{t} u^{2} cos (2 u) d u

\int_{0}^{t} t^{2} cos (2 u) d u = \frac{t ^{2} sin ( 2 t )}{2}

\int_{0}^{t} 2 t u cos (2 u) d u = \frac{t ( 2 t sin ( 2 t ) + cos ( 2 t ) - 1 )}{2}

\int_{0}^{t} u^{2} cos (2 u) d u = \frac{t ^{2} sin ( 2 t ) + t cos ( 2 t )}{2} - \frac{1}{4} sin (2 t)

= \frac{t ^{2} sin ( 2 t )}{2} - \frac{t ( 2 t sin ( 2 t ) + cos ( 2 t ) - 1 )}{2} + \frac{t ^{2} sin ( 2 t ) + t cos ( 2 t )}{2} - \frac{1}{4} sin (2 t)

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