laplacetransform cos(t-pi/6)

解答

拉普拉斯变换

cos (t - \frac{π}{6})

\frac{3 s + 1}{2 ( s ^{2} + 1 )}

求解步骤

L {cos (t - \frac{π}{6})}

使用拉普拉斯变换表:

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

L {cos (t - \frac{π}{6})} = \frac{s cos ( - \frac{π}{6} ) - 1 \cdot sin ( - \frac{π}{6} )}{s ^{2} + 1 ^{2}}

= \frac{s cos ( - \frac{π}{6} ) - 1 \cdot sin ( - \frac{π}{6} )}{s ^{2} + 1 ^{2}}

= \frac{3 s + 1}{2 ( s ^{2} + 1 )}

\frac{5 d u}{d t} = u^{2}

\frac{\partial}{\partial x} (cos (θ))

\int \frac{x ^{6}}{( 8 + x ^{7} ) ^{2}} d x

s l o p e f (x) = 3 x^{4} + 2 x + 8

d er i v a t i v e f (x) = 1 - \frac{1}{x ^{\frac{2}{3}}}