integral from 0 to pi/6 of tsin(2t)

Lösung

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

\frac{- π + 3 3}{24}

Dezimale

0.08560 \dots

Schritte zur Lösung

\int_{0}^{\frac{π}{6}} t sin (2 t) d t

Wende die partielle Integration an

= [\frac{1}{2} (- t cos (2 t) - 2 \cdot \int - \frac{1}{2} cos (2 t) d t)]_{0}^{\frac{π}{6}}

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

= [\frac{1}{2} (- t cos (2 t) - 2 (- \frac{1}{4} sin (2 t)))]_{0}^{\frac{π}{6}}

Vereinfache

[\frac{1}{2} (- t cos (2 t) - 2 (- \frac{1}{4} sin (2 t)))]_{0}^{\frac{π}{6}} : [\frac{1}{2} (- t cos (2 t) + \frac{1}{2} sin (2 t))]_{0}^{\frac{π}{6}}

= [\frac{1}{2} (- t cos (2 t) + \frac{1}{2} sin (2 t))]_{0}^{\frac{π}{6}}

Berechne die Grenzen:

\frac{- π + 3 3}{24}

= \frac{- π + 3 3}{24}

in v erse l a pl a ce \frac{4}{s ^{2} ( s + 2 )} + \frac{3}{s + 2}

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\int \frac{2 x ^{2} + 3}{x ( x - 1 ) ^{2}} d x