Question

Find the Laplace transform of sint.sin2t.sin3t​

Answer 1

Given:

$\[f\left( t \right) = sin\left( t \right).sin\left( {2t} \right).sin\left( {3t} \right)\]$

To Find:

Laplace transformation of given function.

Solution:  

Firstly simplify the given equation-

$\[f\left( t \right) = sin\left( t \right).sin\left( {2t} \right).sin\left( {3t} \right)\]$

$\[f(t) = \sin t(\sin 2t \times \sin 3t)\]$

$\[f(t) = \sin [t][\tfrac{{\cos [2t - 3t]}}{2} - \tfrac{{\cos [2t + 3t]}}{2}]\]$

$\[f(t) = sin[t][\cos [ - t]/2 - \cos [5t]/2]\]$

$\[f(t) = sin[t] \times \frac{1}{2}(\cos [t] - \cos [5t])\]$

$\[f(t) = 1/2(\sin [t]\cos [t] - \sin [t]\cos [5t])\]$

$\[f(t) = 1/2[1/2\sin [2t] - (\sin [ - 4t]/2 + \sin [6t]/2)]\]$

$\[f(t) = 1/2 \times 1/2[\sin [2t] - ( - \sin [4t] + \sin [6t]]\]$

$\[f(t) = 1/4[\sin [2t] + \sin [4t] - \sin [6t]]\]$

Hence, Laplace transformation is-

$\[\begin{gathered} L(f(t)) = L(1/4(\sin 2t + \sin 4t - \sin 6t)) \hfill \\ = L(1/4)L(\sin 2t + \sin 4t - \sin 6t) \hfill \\ = (1/4s)[L(\sin 2t) + L(\sin 4t) - L(\sin 6t)] \hfill \\ = (1/4s)[(\tfrac{2}{{{s^2} + {2^2}}}) + (\tfrac{4}{{{s^2} + {4^2}}}) - (\tfrac{6}{{{s^2} + {6^2}}})] \hfill \\ = (1/4s)[(\tfrac{2}{{{s^2} + 4}}) + (\tfrac{4}{{{s^2} + 16}}) - (\tfrac{6}{{{s^2} + 16}})] \hfill \\ \end{gathered} \]$

Answer 2

Answer:

Step-by-step explanation: