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)

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## 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/(s^{2}+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/(x^{2}+a^{2}) = (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 byL{sin(2t)/t} = tan ^{-1}(2/s). |

**Also Read:**

Laplace transform of 1: | 1/s |

Laplace transform of t: | 1/s^{2} |

Laplace transform of sin t: | 1/(s^{2}+1) |

Laplace transform of cos t: | s/(s^{2}+1) |

## FAQs

**Q1: What is the Laplace of sin2t/t?**

Answer: The Laplace of sin2t/t is equal to tan^{-1}(2/s), that is, L{sin2t/t} = tan^{-1}(2/s).