laplacetransform cos(t)*cos(pi)

解答

拉普拉斯变换

cos (t) \cdot cos (π)

- \frac{s}{s ^{2} + 1}

求解步骤

L {cos (π) cos (t)}

使用三角恒等式改写

L {\frac{1}{2} cos (t + π) + \frac{1}{2} cos (t - π)}

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}{2} L {cos (t + π)} + \frac{1}{2} L {cos (t - π)}

L {cos (t + π)} : - \frac{s}{s ^{2} + 1}

L {cos (t - π)} : - \frac{s}{s ^{2} + 1}

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

整理

\frac{1}{2} (- \frac{s}{s ^{2} + 1}) + \frac{1}{2} (- \frac{s}{s ^{2} + 1}) : - \frac{s}{s ^{2} + 1}

= - \frac{s}{s ^{2} + 1}

\int sin^{5} (3 x) cos (3 x) d x

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