laplacetransform sin(2(t-pi/2))

解答

拉普拉斯变换

sin (2 (t - \frac{π}{2}))

- \frac{2}{s ^{2} + 4}

求解步骤

L {sin (2 (t - \frac{π}{2}))}

使用拉普拉斯变换表:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (2 (t - \frac{π}{2}))} = \frac{s sin ( - π ) + 2 cos ( - π )}{s ^{2} + 2 ^{2}}

= \frac{s sin ( - π ) + 2 cos ( - π )}{s ^{2} + 2 ^{2}}

= - \frac{2}{s ^{2} + 4}

a re a x^{2} - 4 x, x - 4

\int sin (9 π x) d x

\frac{d}{d x} (\frac{3 x ^{4} + 7}{x ^{2}})

\int 2 x (x^{2} + 6)^{6} d x

R \to 1 + lim (\frac{1}{R - 1})