integral from 0 to t of te^{2t}sin(t)

解答

\int_{0}^{t} t e^{2 t} sin (t) d t

\frac{- e ^{2 t} t cos ( t ) + 2 e ^{2 t} t sin ( t )}{5} - \frac{3}{25} e^{2 t} sin (t) + \frac{4}{25} e^{2 t} cos (t) - \frac{4}{25}

求解步骤

\int_{0}^{t} t e^{2 t} sin (t) d t

使用分布积分法

= [t (- \frac{1}{5} e^{2 t} cos (t) + \frac{2}{5} e^{2 t} sin (t)) - \int \frac{1}{5} (- e^{2 t} cos (t) + 2 e^{2 t} sin (t)) d t]_{0}^{t}

\int \frac{1}{5} (- e^{2 t} cos (t) + 2 e^{2 t} sin (t)) d t = \frac{1}{25} (3 e^{2 t} sin (t) - 4 e^{2 t} cos (t))

= [t (- \frac{1}{5} e^{2 t} cos (t) + \frac{2}{5} e^{2 t} sin (t)) - \frac{1}{25} (3 e^{2 t} sin (t) - 4 e^{2 t} cos (t))]_{0}^{t}

带入上下限计算后得:

t (- \frac{1}{5} e^{2 t} cos (t) + \frac{2}{5} e^{2 t} sin (t)) - \frac{1}{25} (3 e^{2 t} sin (t) - 4 e^{2 t} cos (t)) - \frac{4}{25}

= t (- \frac{1}{5} e^{2 t} cos (t) + \frac{2}{5} e^{2 t} sin (t)) - \frac{1}{25} (3 e^{2 t} sin (t) - 4 e^{2 t} cos (t)) - \frac{4}{25}

化简

= \frac{- e ^{2 t} t cos ( t ) + 2 e ^{2 t} t sin ( t )}{5} - \frac{3}{25} e^{2 t} sin (t) + \frac{4}{25} e^{2 t} cos (t) - \frac{4}{25}

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