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

Answer: L{sin5t} = 5/(8s2+8) – 15/(16s2+144) + 5/(16s2+400).

Proof:

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

Step 1:

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

Step 2:

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-e^t)/t

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