laplacetransform sin(2t)cos(3t)

解答

拉普拉斯变换

sin (2 t) cos (3 t)

\frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}

求解步骤

L {sin (2 t) cos (3 t)}

使用三角恒等式改写

L {\frac{1}{2} sin (5 t) - \frac{1}{2} 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}{2} L {sin (5 t)} - \frac{1}{2} L {sin (t)}

L {sin (5 t)} : \frac{5}{s ^{2} + 25}

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

= \frac{1}{2} \cdot \frac{5}{s ^{2} + 25} - \frac{1}{2} \cdot \frac{1}{s ^{2} + 1}

整理

\frac{1}{2} \frac{5}{s ^{2} + 25} - \frac{1}{2} \frac{1}{s ^{2} + 1} : \frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}

= \frac{5}{2 ( s ^{2} + 25 )} - \frac{1}{2 ( s ^{2} + 1 )}

\int 15 (tan (x)) (ln (cos (x))) d x

\int \frac{2 x}{x ^{2} - 4} d x

x \to 0 lim (\frac{( x ^{2} + 2 x )}{x})

t an g e n t y = 4.2 x - x^{2}

\int \frac{700 x}{( x + 1 ) ^{\frac{1}{3}}} d x