integral from 0 to t of cos(x)sin(t-x)

Lösung

\int_{0}^{t} cos (x) sin (t - x) d x

\frac{t sin ( t )}{2}

Schritte zur Lösung

\int_{0}^{t} cos (x) sin (t - x) d x

Wende die partielle Integration an

= [cos (t - x) cos (x) - \int - cos (t - x) sin (x) d x]_{0}^{t}

\int - cos (t - x) sin (x) d x = - \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t))

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

Vereinfache

= [cos (t - x) cos (x) + \frac{1}{4} (sin (t) (2 x - t) - cos (2 x - t))]_{0}^{t}

Berechne die Grenzen:

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

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

Vereinfache

= \frac{t sin ( t )}{2}

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