What is laplacetransform sin(t+(11pi)/2) ?

The answer to laplacetransform sin(t+(11pi)/2) is -s/(s^2+1)

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laplacetransform sin(t+(11pi)/2)

laplace transform

sin (t + \frac{11 π}{2})

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

Solution steps

L {sin (t + \frac{11 π}{2})}

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 + \frac{11 π}{2})} = \frac{s sin ( \frac{11 π}{2} ) + 1 \cdot cos ( \frac{11 π}{2} )}{s ^{2} + 1 ^{2}}

= \frac{s sin ( \frac{11 π}{2} ) + 1 \cdot cos ( \frac{11 π}{2} )}{s ^{2} + 1 ^{2}}

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

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y^{^{'}} = \frac{6 x ^{5} y}{ln ( y )}

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\int_{6}^{8} \frac{68}{( x - 6 ) ^{3}} d x

What is laplacetransform sin(t+(11pi)/2) ?
The answer to laplacetransform sin(t+(11pi)/2) is -s/(s^2+1)