What is laplacetransform cos(t-1) ?

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

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laplacetransform cos(t-1)

laplace transform

cos (t - 1)

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

Solution steps

L {cos (t - 1)}

Use Laplace Transform table:

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

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

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

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

\int \frac{1}{x ^{2} - 324} d x

\frac{d}{d x} ((x^{2} - 6) e^{8 x})

\int 0 x d x

y = x^{2} (5 x - 1)^{3}

f (x) = (3 x + 1) (5 x + 4)

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