find Laplace transform t^3 . cos 4t


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Question Number 88169 by jagoll last updated on 08/Apr/20

$find Laplace transform$ $t^{3} . \cos 4 t$
Commented by mathmax by abdo last updated on 08/Apr/20
$L(x^3 cos(4x))=∫_0 ^∞ f(t) e^(−xt) dt =∫_0 ^∞ t^3 cos(4t)e^(−xt) dt =Re(∫_0 ^∞ t^3 e^(i4t−xt) dt) ∫_0 ^∞ t^3 e^((−x+4i)t) dt =_(bypsrts) [(t^3 /(−x+4i)) e^((−x+4i)t) ]_0 ^∞ −(1/(−x+4i))∫_0 ^∞ 3t^2 e^((−x+4i)t) dt =(3/(x−4i)) { [(t^2 /(−x+4i)) e^((−x+4i)t) ]_0 ^∞ −(1/(−x+4i))∫_0 ^∞ 2t e^((−x+4i)t) dt} =(6/((x−4i)^2 )) ∫_0 ^∞ t e^((−x+4i)t) dt =(6/((x−4i)^2 )){ [−(t/(−x+4i)) e^((−x+4i)t) ]_0 ^∞ −(1/(−x+4i))∫_0 ^∞ e^((−x+4i)t) dt} =(6/((x−4i)^3 ))[(1/(−x+4i)) e^((−x+4i)t) ]_0 ^∞ =(6/((x−4i)^4 )) =6((1/(x−4i)))^4 =6(((x+4i)/(x^2 +16)))^4 =(6/(x^2 +16))(x+4i)^2 (x+4i)^2 =(6/(x^2 +16))(x^2 +8ix −16)^2 =(6/(x^2 +16)){ (x^2 +8ix)^2 −32(x^2 +8ix)+16^2 } =(6/(x^2 +16)){ x^4 +16ix^3 −64 x^2 −32x^2 −8.32ix +16^2 } ⇒ L(x^3 (4x)) =(6/(x^2 +16))( x^4 −96x^2 +16^2 )$
$L (x^{3} c o s (4 x)) = \int_{0}^{\infty} f (t) e^{- x t} d t$ $= \int_{0}^{\infty} t^{3} c o s (4 t) e^{- x t} d t = R e (\int_{0}^{\infty} t^{3} e^{i 4 t - x t} d t)$ $\int_{0}^{\infty} t^{3} e^{(- x + 4 i) t} d t =_{b y p s r t s} {[\frac{t^{3}}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} 3 t^{2} e^{(- x + 4 i) t} d t$ $= \frac{3}{x - 4 i} {{[\frac{t^{2}}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} 2 t e^{(- x + 4 i) t} d t}$ $= \frac{6}{{(x - 4 i)}^{2}} \int_{0}^{\infty} t e^{(- x + 4 i) t} d t$ $= \frac{6}{{(x - 4 i)}^{2}} {{[- \frac{t}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} - \frac{1}{- x + 4 i} \int_{0}^{\infty} e^{(- x + 4 i) t} d t}$ $= \frac{6}{{(x - 4 i)}^{3}} {[\frac{1}{- x + 4 i} e^{(- x + 4 i) t}]}_{0}^{\infty} = \frac{6}{{(x - 4 i)}^{4}} = 6 {(\frac{1}{x - 4 i})}^{4}$ $= 6 {(\frac{x + 4 i}{x^{2} + 16})}^{4} = \frac{6}{x^{2} + 16} {(x + 4 i)}^{2} {(x + 4 i)}^{2}$ $= \frac{6}{x^{2} + 16} {(x^{2} + 8 i x - 16)}^{2}$ $= \frac{6}{x^{2} + 16} {{(x^{2} + 8 i x)}^{2} - 32 (x^{2} + 8 i x) + 16^{2}}$ $= \frac{6}{x^{2} + 16} {x^{4} + 16 {i x}^{3} - 64 x^{2} - 32 x^{2} - 8.32 i x + 16^{2}} \Rightarrow$ $L (x^{3} (4 x)) = \frac{6}{x^{2} + 16} (x^{4} - 96 x^{2} + 16^{2})$
Commented by mathmax by abdo last updated on 09/Apr/20

$f o r g i v e L (x^{3} c o s (4 x)) = \frac{6 (x^{4} - 96 x^{2} + 16^{2})}{{(x^{2} + 16)}^{4}}$
	Answered by jagoll last updated on 09/Apr/20

	$L (\cos 4 t) = \frac{s}{s^{2} + 16}$ $L (t \cos 4 t) = - \frac{d}{ds} [\frac{s}{s^{2} + 16}] = \frac{16 - s^{2}}{{(s^{2} + 16)}^{2}}$ $L (t^{2} \cos 4 t) = - \frac{d}{ds} [\frac{16 - s^{2}}{{(s^{2} + 16)}^{2}}]$ $= \frac{96 s - {2 s}^{3}}{{(s^{2} + 16)}^{3}}$ $L (t^{3} \cos 4 t) = - \frac{d}{ds} [\frac{96 s - {2 s}^{3}}{{(s^{2} + 16)}^{3}}]$ $= \frac{6 (s^{4} - {96 s}^{2} + 256)}{{(s^{2} + 16)}^{4}}$
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