The Laplace transform of sin^4t is denoted by L{sin^{4}t}, and it is equal to 3/(8s) – s/[2(s^{2}+4)] + s/[8(s^{2}+16)]. So the Laplace formula of sin^{4}t is equal to

L{sin^{4}t} = $\dfrac{3}{8s} – \dfrac{s}{2(s^2+4)} + \dfrac{s}{8(s^2+16)}$.

Let us now prove the above Laplace transform formula of the fourth power of sint.

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## Laplace of sin^{4}t

**Question:** What is the Laplace of sin^{4}t?

Answer: L{sin^{4}t} = 3/(8s) – s/[2(s^{2}+4)] + s/[8(s^{2}+16)]. |

**Proof:**

To find the Laplace transform of sin^{4}t, we will follow the below steps.

**Step 1:**

First, we simplify the function sin^{4}t using the trigonometric identity: 2sin^{2}t = 1-cos2t. Therefore,

sin^{4}t = sin^{2}t × sin^{2}t

⇒ sin^{4}t = $\big( \dfrac{1-\cos 2t}{2}\big)^2$

⇒ sin^{4}t = $\dfrac{1-2\cos 2t+\cos^2 2t}{4}$

⇒ sin^{4}t = $\dfrac{1-2\cos 2t+ \frac{1+\cos 4t}{2}}{4}$ by the identity 2cos^{2}θ = 1+cos2θ.

⇒ sin^{4}t = $\dfrac{2-4\cos 2t+ 1+\cos 4t}{8}$

⇒ sin^{4}t = $\dfrac{3-4\cos 2t+\cos 4t}{8}$.

**Step 2:**

So the Laplace of sin^{4}t will be equal to

L{sin^{4}t} = L$\big\{ \dfrac{3-4\cos 2t+\cos 4t}{8} \big\}$

= $\dfrac{1}{8} \big[ 3L\{1\} – 4L\{\cos 2t\}+ L\{\cos 4t\} \big]$, by the property of Laplace transforms.

= $\dfrac{1}{8} \big[ \dfrac{3}{s} – 4 \dfrac{s}{s^2+2^2}+ \dfrac{s}{s^2+4^2} \big]$ as we know that L{1} = 1/s and L{cosat} = s/(s^{2}+a^{2}).

= $\dfrac{3}{8s} – \dfrac{s}{2(s^2+4)}+ \dfrac{s}{8(s^2+16)}$.

Thus, the Laplace transform of sin^{4}t is equal to L{sin^{4}t} = 3/(8s) – s/[2(s^{2}+4)] + s/[8(s^{2}+16)].

Also Read:

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

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

## FAQs

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

^{4}t?Answer: The Laplace transform of sin^{4}t is L{sin^{4}t} = 3/(8s) – s/(2s^{2}+8) + s/(8s^{2}+128).