laplacetransform sin^3(t)

解答

拉普拉斯变换

sin^{3} (t)

\frac{6}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

求解步骤

L {sin^{3} (t)}

使用三角恒等式改写

L {- \frac{1}{4} sin (3 t) + \frac{3}{4} sin (t)}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= - \frac{1}{4} L {sin (3 t)} + \frac{3}{4} L {sin (t)}

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

L {sin (t)} : \frac{1}{s ^{2} + 1}

= - \frac{1}{4} \cdot \frac{3}{s ^{2} + 9} + \frac{3}{4} \cdot \frac{1}{s ^{2} + 1}

化简

- \frac{1}{4} \frac{3}{s ^{2} + 9} + \frac{3}{4} \frac{1}{s ^{2} + 1} : \frac{6}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

= \frac{6}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

\int x 5^{- x^{2}} d x

d er i v a t i v e sec (x)

in v erse l a pl a ce \frac{1}{( s - 2 ) ( s + 1 )}

s l o p e in t erce pt (6, - 2), (12, 1)

x \to 2 lim (x^{2} + 3 x)