What is laplacetransform sin(2t+5) ?

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

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laplacetransform sin(2t+5)

laplace transform

sin (2 t + 5)

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

Solution steps

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

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

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

\int \frac{ln ^{2} ( x )}{x ^{\frac{5}{3}}} d x

x \to \infty lim (∣ 1 ∣)

\int_{0}^{t} e^{u} d u

x \to \infty lim (4^{x})

\int_{- 1}^{2} (7 - 3 x) d x

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