What is laplacetransform sin(t-1) ?

The answer to laplacetransform sin(t-1) is (-sin(1)s+cos(1))/(s^2+1)

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laplacetransform sin(t-1)

laplace transform

sin (t - 1)

\frac{- sin ( 1 ) s + cos ( 1 )}{s ^{2} + 1}

Solution steps

L {sin (t - 1)}

Use Laplace Transform table:

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

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

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

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

\int x^{2} (5 x^{3} + 7 x^{2} - 6 x - 3) d x

sin (2 x) d x + cos (3 y) d y = 0, y (\frac{π}{2}) = \frac{π}{3}

\int \frac{4}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

x \to + 3 lim (4 - x)

x \to 0 lim (x + x)

What is laplacetransform sin(t-1) ?
The answer to laplacetransform sin(t-1) is (-sin(1)s+cos(1))/(s^2+1)