laplacetransform sin^2(2t)

解答

拉普拉斯变换

sin^{2} (2 t)

\frac{8}{s ( s ^{2} + 16 )}

求解步骤

L {sin^{2} (2 t)}

使用三角恒等式改写

L {\frac{1}{2} - \frac{1}{2} cos (4 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)}

= L {\frac{1}{2}} - \frac{1}{2} L {cos (4 t)}

L {\frac{1}{2}} : \frac{1}{2 s}

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 16}

化简

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 16} : \frac{8}{s ( s ^{2} + 16 )}

= \frac{8}{s ( s ^{2} + 16 )}

\int \frac{4 x ^{3}}{1 - x ^{4}} d x

\int (4 x^{2} + 3)^{2} d x

\frac{\partial}{\partial x} (5 y - 2 x)

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

d er i v a t i v e 5 cos^{2} (x) + 5 sin (x)