Find Laplace transform of sin2t sin3t

The Laplace transform of sin2t sin3t is equal to 12s/(s²+1)(s²+25). Here we learn how to find the Laplace of sin2t sin3t.

Note that sin2t sin3t is the product of two sine functions. The Laplace transform formula of sin2t sin3t is given below:

L{sin2t sin 3t} = $\dfrac{12s}{(s^2+1)(s^2+25)}$.

Laplace of sin2t sin3t

Answer: The Laplace of sin2t sin3t is 12s/(s2+1)(s2+25).

Proof:

To find the Laplace of sin2t sin3t, we will use the formula of sinx siny which is given below:

sinx siny = $\dfrac{1}{2} [\cos(x-y)-\cos(x+y)]$.

So we have that

sin2t sin3t = $\dfrac{1}{2} [\cos(2t-3t)-\cos(2t+3t)]$

⇒ sin2t sin3t = $\dfrac{1}{2} (\cos t-\cos 5t)$ as cos(-x)=cosx.

Therefore,

L{sin2t sin3t} = L $\big\{\dfrac{1}{2} (\cos t-\cos 5t) \big\}$

= $\dfrac{1}{2} L\{ \cos t-\cos 5t\}$

= $\dfrac{1}{2} L\{ \cos t \}$ – $\dfrac{1}{2} L\{ \cos 5t \}$ by the linear property of Laplace transforms.

= $\dfrac{1}{2} \dfrac{s}{s^2+1}$ – $\dfrac{1}{2} \dfrac{s}{s^2+25}$, here we have used the formula of L{cos at} = s/(s²+a²).

= $\dfrac{s}{2} \dfrac{s^2+25-s^2-1}{(s^2+1)(s^2+25)}$

$\dfrac{12s}{(s^2+1)(s^2+25)}$.

So the Laplace of transform of sin2t sin3t is 12s/(s²+1)(s²+25).

Main Article: 

Laplace Transform: Definition, Table, Formulas, Properties

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