# Laplace Transform of sin^5t | Find L{sin^5t}

The Laplace transform of sin^5t is equal to 5/(8s2+8) – 15/(16s2+144) + 5/(16s2+400). So the Laplace formula of sin5t is given as follows.

L{sin5t} = $\dfrac{5}{8s^2+8} – \dfrac{15}{16s^2+144} + \dfrac{5}{16s^2+400}$.

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

## What is the Laplace of sin5t

To find the Laplace transform of sin5t, we need to follow the below steps.

First, we simplify the function sin5t using the following trigonometric identities:

1. 2sin2t = 1-cos2t.
2. 4sin3t = 3sint -sin3t.

So sin5t is reduced as follows.

sin5t = sin2t × sin3t

⇒ sin5t = $\dfrac{1-\cos 2t}{2} \times \dfrac{3\sin t -\sin 3t}{4}$

⇒ sin5t = $\dfrac{1}{8}$ (3sint – sin3t -3sint cos2t +sin3t cos2t)

⇒ sin5t = $\dfrac{1}{8} (3\sin t – \sin 3t – \dfrac{3\sin 3t -3\sin t}{2} + \dfrac{\sin 5t + \sin t}{2})$. This is because sinA cosB = [sin(A+B) + sin(A-B)]/2.

⇒ sin5t = $\dfrac{1}{16} (6\sin t – 2\sin 3t – 3\sin 3t +3\sin t + \sin 5t + \sin t)$

⇒ sin5t = $\dfrac{1}{16} (10\sin t – 5\sin 3t + \sin 5t)$.

Now, taking Laplace transforms on both sides we obtain that

L{sin5t} = 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/(s2+a2).

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

Thus, the Laplace transform of sin5t is equal to L{sin5t} = 5/(8s2+8) – 15/(16s2+144) + 5/(16s2+400).

More Laplace Transforms:

Laplace transform of sin3t

Laplace transform of sin4t

Laplace transform of cosht

Laplace transform of u(t-2)

Laplace transform of t2u(t-1)

Laplace transform of (1-et)/t

## FAQs

Q1: What is the Laplace transform of sin5t?

Answer: The Laplace transform of sin5t is L{sin5t} = 5/[8(s2+1)] – 15/[16(s2+9)] + 5/[16(s2+25)].

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