The Laplace transform of sin2t/t is equal to tan-1(2/s). In this article, we will learn how to find the Laplace of sin2t/t. The Laplace transform formula of sin2t/t is given as follows:
L{sin2t/t} = tan-1(2/s).
To evaluate the Laplace trasform of sin2t divided by t, we will use the two formulas listed below:
- L{sinat} = $\dfrac{a}{s^2+a^2}$
- $L\{\frac{f(t)}{t} \} =\int_s^\infty F(s) ds$, where L{f(t)}=F(s)
Table of Contents
What is the Laplace Transform of sin2t/t?
Answer: The Laplace transform of sin(2t)/t is tan-1(2/s).
Proof:
Let us put f(t) = sin2t in the above formula 2. Note that
F(s) = L{f(t)} = L{sin 2t} = 2/(s2+4)
Thus by formula 2, the Laplace transform of sin2t/t is equal to
L{sin2t/t} = $\int_s^\infty \dfrac{2}{s^2+4} ds$
= $\Big[ \tan^{-1} \frac{s}{2}\Big]_s^\infty$ as we know that ∫dx/(x2+a2) = (1/a) tan-1(x/a).
= [tan-1 ∞ – tan-1 (s/2)]
= [π/2 – tan-1(s/2)]
= cot-1(s/2)
= tan-1 (2/s).
Therefore, the Laplace transform of sin2t/t is tan-1(2/s).
| Summary: The Laplace transform of sin2t/t is given by L{sin(2t)/t} = tan-1(2/s). |
Also Read:
| Laplace transform of 1: | 1/s |
| Laplace transform of t: | 1/s2 |
| Laplace transform of sin t: | 1/(s2+1) |
| Laplace transform of cos t: | s/(s2+1) |
FAQs
Answer: The Laplace of sin2t/t is equal to tan-1(2/s), that is, L{sin2t/t} = tan-1(2/s).
