What is laplacetransform sin(t+c) ?

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

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laplacetransform sin(t+c)

laplace transform

sin (t + c)

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

Solution steps

L {sin (t + c)}

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

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

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

\int cos^{4} (\frac{x}{5}) d x

\int \frac{x + 3}{x ^{2} 2 x + 3} d x

y^{^{'}} = \frac{x ^{4} y ^{4} - 2 x ^{4}}{y ^{3}}

x \to \infty lim (3 x^{2} + 1)

\frac{d}{d x} (\frac{2}{3} x^{3} + x^{2} - 12 x + 17)

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