The Laplace transform of t sin2t is equal to L{t sin2t}= 4s/(s^{2}+4)^{2}. In this post, we learn how to find the Laplace of t times sin2t.

The Laplace formula of tsin2t is given by

L{t sin2t}= $\dfrac{4s}{(s^2+4)^2}$.

Table of Contents

## Find Laplace of t sin2t

**Question:** What is the Laplace of tsin2t?

**Solution:**

If L{f(t)}=F(s), then the multiplication by t Laplace formula states that

L{t^{n} f(t)} = (-1)^{n} $\dfrac{d^n}{ds^n} \big(F(s) \big)$ **…(I)**

Put n=1 and f(t) = sin2t.

So F(s) = L{sin2t} = $\dfrac{2}{s^2+4}$ as we know L{sin at} = a/(s^{2}+a^{2}).

By the above formula (I),

L{t sin2t} = (-1)^{1} $\dfrac{d}{ds} \Big( \dfrac{2}{s^2+4} \Big)$

= $-2\dfrac{d}{ds} (s^2+4)^{-1}$

= -2 × $[-1(s^2+4)^{-2} \frac{d}{ds}(s^2+4)]$ by the power rule of derivatives.

= -2 × $-2s(s^2+4)^{-2}$

= $\dfrac{4s}{(s^2+4)^2}$

So the Laplace transform of tsin2t is 4s/(s^{2}+4)^{2}.

More Laplace Transforms:

Laplace transform of sin2t sin3t

Laplace transform of sint sin2t sin3t

## FAQs

**Q1: What is the Laplace transform of tsin2t?**

Answer: The Laplace transform of tsin2t is equal to 4s/(s^{2}+4)^{2}.

**Q2: Find L{t sin2t}.**

Answer: L{t sin2t}= 4s/(s^{2}+4)^{2}.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.