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_{0}^{4} x^{2} - 4 x d x

\frac{1}{t + 1}

\int t e^{- 5 t} d t

x \to 5 lim (x^{2} - 8 x + 15)

x \to 0 lim (\frac{2}{x})

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