Laplace transform of sint sin2t sin3t

The Laplace transform of sint sin2t sin3t is equal to 1/(s²+16)+1/{2(s²+4)} -3/{2(s²+36)}. The Laplace formula of sin t sin 2t sin 3t is given below:

L{sint sin2t sin3t} = $\dfrac{1}{s^2+16}+\dfrac{1}{2(s^2+4)}-\dfrac{3}{2(s^2+36)}$.

L{sint sin2t sin3t}

Question: Find the Laplace Transform of sint sin2t sin3t.

Solution:

First, we will simplify the expression sint sin2t sin3t. Note that we have:

sint sin2t sin3t

= $\dfrac{1}{2}$ (2 sint sin2t) sin3t

= $\dfrac{1}{2}$ [cos(t-2t) – cos(t+2t)] sin3t by the formula 2sinx siny = cos(x-y)-cos(x+y).

= $\dfrac{1}{2}$ (cost – cos3t) sin3t as cos(-x)=cosx.

= $\dfrac{1}{4}$ (2cost sin3t – 2cos3t sin3t)

= $\dfrac{1}{4}$ [sin(t+3t) -sin(t-3t) – sin6t] using the formula 2sinx cosx = sin2x, and 2cosx siny = sin(x+y)-sin(x-y).

= $\dfrac{1}{4}$ (sin4t +sin2t – sin6t) as sin(-x)=-sinx.

Thus, we have shown that

sint sin2t sin3t = $\dfrac{1}{4}$ (sin4t +sin2t – sin6t).

Taking Laplace transform on both sides and using the formula L{sin at}=a/(s²+a²), we obtain that

L{sint sin2t sin3t} = L{$\dfrac{1}{4}$ (sin4t +sin2t – sin6t)}

= $\dfrac{1}{4}$ [L{sin4t} + L{sin2t} – L{sin6t}]

= $\dfrac{1}{s^2+16}+\dfrac{1}{2(s^2+4)}-\dfrac{3}{2(s^2+36)}$

So the Laplace transform of sint sin2t sin3t is equal to $\dfrac{1}{s^2+16}+\dfrac{1}{2(s^2+4)}-\dfrac{3}{2(s^2+36)}$.

Related Laplace Transforms:

Laplace Transform of sin2t sin3t

Laplace Transform of sin²t

Laplace Transform of cos²t

Laplace Transform of sin2t/t

Laplace Transform of t sint

FAQs