# Laplace Transform of sin2t/t

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:

1. L{sinat} = $\dfrac{a}{s^2+a^2}$
2. $L\{\frac{f(t)}{t} \} =\int_s^\infty F(s) ds$, where L{f(t)}=F(s)

## 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).