What is laplacetransform sin(t+2pi) ?

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

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laplacetransform sin(t+2pi)

laplace transform

sin (t + 2 π)

\frac{1}{s ^{2} + 1}

Solution steps

L {sin (t + 2 π)}

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

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

= \frac{1}{s ^{2} + 1}

\int (2 sin (2 x) - 2 cos (2 x)) d x

\int x^{4} sin (4 x) d x

x \to 1 lim ((x - 1)^{(x + 4)})

(2 x y^{2} - 6) d x + (2 x^{2} y + 7) d y = 0

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

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