展开 (e^{(-st-2t)})(cos(2t)-4sin(2t))dt

解答

展开

(e^{(- s t - 2 t)}) (cos (2 t) - 4 sin (2 t)) d t

e^{- s t - 2 t} cos (2 t) d t - 4 e^{- s t - 2 t} sin (2 t) d t

求解步骤

(e^{(- s t - 2 t)}) (cos (2 t) - 4 sin (2 t)) d t

使用分配律:

a (b - c) = ab - a c

a = (e^{(- s t - 2 t)}) d t, b = cos (2 t), c = 4 sin (2 t)

= (e^{(- s t - 2 t)}) d t cos (2 t) - (e^{(- s t - 2 t)}) d t \cdot 4 sin (2 t)

= cos (2 t) (e^{(- s t - 2 t)}) d t - 4 sin (2 t) (e^{(- s t - 2 t)}) d t

去除括号:

(a) = a

= e^{- s t - 2 t} cos (2 t) d t - 4 e^{- s t - 2 t} sin (2 t) d t

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e x p an d (5 x + \frac{3}{2} y + z)^{2}

e x p an d (x^{2} - x sec^{2} (y)) d y

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e x p an d (8 t^{7}, 5 t^{3}, - 4 t^{3}) l e f t (1, 2 t, 3 t^{2})