# Laplace Transform of t sinht | Find L{tsinh at}

The Laplace Transform of t sinht is equal to 2s/(s2-1)2. The Laplace of the product tsinh t is denoted by L{t sinht}, and its formula is given as follows:

$\boxed{L\{t \sinh t\} = \dfrac{2s}{(s^2-1)^2}}$

More generally, L{t sinhat} = 2as/(s2-a2)2. In this post, we will find the Laplace transform of tsinh(at).

## Laplace of t sinht

Answer: The Laplace of tsinht is L{t sinht} = 2s/(s2-1)2.

By the multiplication by t Laplace transform formula, we will find the Laplace transform of t sinht. If L{f(t)}=F(s), then this formula states that

$L\{t f(t)\} = – \dfrac{d}{ds}(F(s))$ …(∗)

Step 1: Put f(t) = sinht

∴ F(s) = L{f(t)} = L{sinht} = 1/(s2-1) by the Laplace of sinh(at).

Step 2: Now using (∗) the Laplace transform of tsinh(t) is

$L\{t\sinh t\} = – \dfrac{d}{ds}\left(\dfrac{1}{s^2-1}\right)$

Step 3: By the quotient rule of derivatives, it follows that

$L\{t\sinh t \}$ $= – \dfrac{(s^2-1)\frac{d}{ds}(1)-1 \frac{d}{ds}(s^2-1)}{(s^2-1)^2}$

$= – \dfrac{(s^2-1)\cdot 0- 1\cdot 2s}{(s^2-1)^2}$

$= – \dfrac{-2s}{(s^2-1)^2}$

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

So the Laplace transform of tsinh t is equal to 2s/(s2-1)2.

More Laplace Transforms:

## What is the Laplace Transform of t sinh(at)?

Proof:

As L{sinh at} = a/(s2-a2), using the above formula (∗), the Laplace of tsinh(at) is equal to

$L\{t\sinh(at)\} = – \dfrac{d}{ds}\left(\dfrac{a}{s^2-a^2}\right)$

= $- \dfrac{(s^2-a^2)\frac{d}{ds}(a)-a \frac{d}{ds}(s^2-a^2)}{(s^2-a^2)^2}$

= $- \dfrac{(s^2-a^2)\cdot 0-a \cdot 2s}{(s^2-a^2)^2}$

= $- \dfrac{-2as}{(s^2-a^2)^2}$

= $\dfrac{2as}{(s^2-a^2)^2}$.

Therefore, the Laplace transform of tsinh(at) is 2as/(s2-a2)2.

Laplace transform of (1-sint)/t

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## FAQs

Q1: What is the Laplace transform of t sinht?

Answer: The Laplace transform of t sinht is 2s/(s2-1)2, that is, L{t sinht} = 2s/(s2-1)2.

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