integral from 0 to t of 3ue^{2u}cos(u)

解答

\int_{0}^{t} 3 u e^{2 u} cos (u) d u

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

求解步骤

\int_{0}^{t} 3 u e^{2 u} cos (u) d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{t} cos (u) e^{2 u} u d u

使用分布积分法

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

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

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

带入上下限计算后得:

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

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

化简

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

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