展开 (te^{(-2t)}+sin(t)+cos(t))^2U(t+6)

解答

展开

(t e^{(- 2 t)} + sin (t) + cos (t))^{2} U (t + 6)

e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)

求解步骤

(t e^{(- 2 t)} + sin (t) + cos (t))^{2} U (t + 6)

= U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} (t + 6)

使用分配律:

a (b + c) = ab + a c

a = (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U, b = t, c = 6

= (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U t + (t e^{(- 2 t)} + sin (t) + cos (t))^{2} U \cdot 6

= t U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} + 6 U (e^{(- 2 t)} t + sin (t) + cos (t))^{2}

化简

t U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} + 6 U (e^{(- 2 t)} t + sin (t) + cos (t))^{2} : e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)

= e^{- 4 t} t^{3} U + 2 e^{- 2 t} t^{2} U sin (t) + 2 e^{- 2 t} t^{2} U cos (t) + t U sin^{2} (t) + 2 t U cos (t) sin (t) + t U cos^{2} (t) + 6 e^{- 4 t} t^{2} U + 12 e^{- 2 t} t U sin (t) + 12 e^{- 2 t} t U cos (t) + 6 U sin^{2} (t) + 12 U cos (t) sin (t) + 6 U cos^{2} (t)

e x p an d (6 - 8.5 sin (x))^{2}

s im pl i f y y^{4} (\frac{2 x ^{2}}{y ^{2}} + 1)

e x p an d (2 x + y) d x + (2 y e^{\frac{y}{x}} - x) d y

s im pl i f y (x + i x^{2})^{2} - 2 (x + i x^{2}) (1 + 2 i x)

e x p an d \frac{2 - e ^{- 3 π s}}{s ( s ^{2} + 1 )}