laplacetransform sin(6(t-pi/6))

解答

拉普拉斯变换

sin (6 (t - \frac{π}{6}))

- \frac{6}{s ^{2} + 36}

求解步骤

L {sin (6 (t - \frac{π}{6}))}

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 (6 (t - \frac{π}{6}))} = \frac{s sin ( - π ) + 6 cos ( - π )}{s ^{2} + 6 ^{2}}

= \frac{s sin ( - π ) + 6 cos ( - π )}{s ^{2} + 6 ^{2}}

= - \frac{6}{s ^{2} + 36}

n = 0 \sum \infty \frac{1}{( 4 n + 3 ) ^{2}}

\frac{d}{d x} ((x + 3)^{\frac{1}{2}})

\int \frac{1}{x ^{2} ( 4 + x ^{2} )} d x

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

d er i v a t i v e 4 cos (x) + 12 x