The Laplace transform of cos^2t is equal to 1/2s + s/(2s^{2}+8). In this article, we will learn how to find the Laplace of cos square t. The formula of the Laplace of cos^{2}t is given below:

L{cos^{2}t} = 1/2s + s/(2s^{2}+8).

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## Find the Laplace Transform of cos^{2}t

**Answer:** The Laplace of cos square t is 1/2s + s/(2s^{2}+8).

*Proof:*

From the trigonometric formulas, we know that 1+cos2t=2cos^{2}t. Thus, it follows that

cos^{2}t = $\frac{1}{2}$(1+cos2t)

Taking Laplace transforms on both sides, we get that L{cos^{2}t} = L{$\frac{1}{2}$(1+cos2t)}. Now, by the linearity property of Laplace transforms it follows that

L{cos^{2}t} = $\frac{1}{2}$ (L{1} + L{cos2t}).

As the Laplace Transform of 1 is 1/s and the Laplace transform of cosat is s/(s^{2}+a^{2}), we deduce that

L{cos^{2}t} = $\dfrac{1}{2} \left(\dfrac{1}{s}+\dfrac{s}{s^2+4} \right)$

Simplifying the above, we get that

L{cos^{2}t} = $\dfrac{1}{2s}+\dfrac{s}{2(s^2+4)}$

So the Laplace transform of cos^2t is equal to 1/2s + s/(2s^{2}+8).

**Main Article:** Laplace Transform: Definition, Table, Formulas, Properties

Laplace transform of sin at |

Laplace transform of sint/t |

Laplace transform of sin2t/t |

Laplace transform of cos2t/t |

Laplace transform of t sint |

## FAQs

**Q1: What is the Laplace Transform of cos^2t?**

Answer: The Laplace transform of cos^2t is equal to 1/2s + s/(2s^{2}+8). That is, L{cos^{2}t} = 1/2s + s/(2s^{2}+8).