The Laplace transform of sin^5t is equal to 5/(8s^{2}+8) – 15/(16s^{2}+144) + 5/(16s^{2}+400). So the Laplace formula of sin^{5}t is given as follows.

L{sin^{5}t} = $\dfrac{5}{8s^2+8} – \dfrac{15}{16s^2+144} + \dfrac{5}{16s^2+400}$.

We now find the Laplace transform of the fifth power of sint.

Table of Contents

## What is the Laplace of sin^{5}t

Answer: L{sin^{5}t} = 5/(8s^{2}+8) – 15/(16s^{2}+144) + 5/(16s^{2}+400). |

**Proof:**

To find the Laplace transform of sin^{5}t, we need to follow the below steps.

**Step 1:**

First, we simplify the function sin^{5}t using the following trigonometric identities:

- 2sin
^{2}t = 1-cos2t. - 4sin
^{3}t = 3sint -sin3t.

So sin^{5}t is reduced as follows.

sin^{5}t = sin^{2}t × sin^{3}t

⇒ sin^{5}t = $\dfrac{1-\cos 2t}{2} \times \dfrac{3\sin t -\sin 3t}{4}$

⇒ sin^{5}t = $\dfrac{1}{8}$ (3sint – sin3t -3sint cos2t +sin3t cos2t)

⇒ sin^{5}t = $\dfrac{1}{8} (3\sin t – \sin 3t – \dfrac{3\sin 3t -3\sin t}{2} + \dfrac{\sin 5t + \sin t}{2})$. This is because sinA cosB = [sin(A+B) + sin(A-B)]/2.

⇒ sin^{5}t = $\dfrac{1}{16} (6\sin t – 2\sin 3t – 3\sin 3t +3\sin t + \sin 5t + \sin t)$

⇒ sin^{5}t = $\dfrac{1}{16} (10\sin t – 5\sin 3t + \sin 5t)$.

**Step 2:**

Now, taking Laplace transforms on both sides we obtain that

L{sin^{5}t} = L$\big\{ \dfrac{1}{16} (10\sin t – 5\sin 3t + \sin 5t) \big\}$

= $\dfrac{10}{16} L\{\sin t\} – \dfrac{5}{16} L\{\sin 3t\} + \dfrac{1}{16} L\{\sin 5t\}$, by the Linearity Property of Laplace Transforms.

= $\dfrac{5}{8} \dfrac{1}{s^2+1^2} – \dfrac{5}{16} \dfrac{3}{s^2+3^2}+ \dfrac{1}{16} \dfrac{5}{s^2+5^2}$, using the Laplace transform formula L{sin at} = a/(s^{2}+a^{2}).

= $\dfrac{5}{8(s^2+1)} – \dfrac{15}{16(s^2+9)}+ \dfrac{5}{16(s^2+25)}$.

Thus, the Laplace transform of sin^{5}t is equal to L{sin^{5}t} = 5/(8s^{2}+8) – 15/(16s^{2}+144) + 5/(16s^{2}+400).

More Laplace Transforms:

Laplace transform of t^{2}u(t-1)

Laplace transform of (1-e^{t})/t

## FAQs

**Q1: What is the Laplace transform of sin**

^{5}t?Answer: The Laplace transform of sin^{5}t is L{sin^{5}t} = 5/[8(s^{2}+1)] – 15/[16(s^{2}+9)] + 5/[16(s^{2}+25)].

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.