What is laplacetransform sin(4t+5) ?

The answer to laplacetransform sin(4t+5) is (sin(5)s+4cos(5))/(s^2+16)

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laplacetransform sin(4t+5)

laplace transform

sin (4 t + 5)

\frac{sin ( 5 ) s + 4 cos ( 5 )}{s ^{2} + 16}

Solution steps

L {sin (4 t + 5)}

Use Laplace Transform table:

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

L {sin (4 t + 5)} = \frac{s sin ( 5 ) + 4 cos ( 5 )}{s ^{2} + 4 ^{2}}

= \frac{s sin ( 5 ) + 4 cos ( 5 )}{s ^{2} + 4 ^{2}}

= \frac{sin ( 5 ) s + 4 cos ( 5 )}{s ^{2} + 16}

f (x) = - x^{3}, (- 2, 8)

\int cos (x^{2}) 2 x d x

n = 1 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

\int \frac{e ^{5 x}}{e ^{5 x} + 6} d x

- 2 x y - 4 x^{4} = - 2 + y

What is laplacetransform sin(4t+5) ?
The answer to laplacetransform sin(4t+5) is (sin(5)s+4cos(5))/(s^2+16)