Shûbhâm Pątïĺ

2022-07-16

$L\left[{\mathrm{cos}}^{2}x\right]$

xleb123

To solve the given problem, which is finding the Laplace transform of ${\mathrm{cos}}^{2}\left(x\right)$, we'll apply the definition and properties of the Laplace transform.
The Laplace transform of a function $f\left(t\right)$ is defined as:
$ℒ\left\{f\left(t\right)\right\}=F\left(s\right)={\int }_{0}^{\infty }{e}^{-st}f\left(t\right)dt$
In this case, we want to find the Laplace transform of ${\mathrm{cos}}^{2}\left(x\right)$. However, the Laplace transform is typically used for functions defined on the interval $\left[0,\infty \right)$. The function ${\mathrm{cos}}^{2}\left(x\right)$ is defined for all real values of $x$. To proceed, we need to express ${\mathrm{cos}}^{2}\left(x\right)$ in terms of a function defined on the interval $\left[0,\infty \right)$.
Using the identity ${\mathrm{cos}}^{2}\left(x\right)=\frac{1}{2}\left(1+\mathrm{cos}\left(2x\right)\right)$, we can rewrite ${\mathrm{cos}}^{2}\left(x\right)$ as $\frac{1}{2}\left(1+\mathrm{cos}\left(2x\right)\right)$.
Now, we can find the Laplace transform of ${\mathrm{cos}}^{2}\left(x\right)$ by applying the linearity property of the Laplace transform.
$ℒ\left\{{\mathrm{cos}}^{2}\left(x\right)\right\}=ℒ\left\{\frac{1}{2}\left(1+\mathrm{cos}\left(2x\right)\right)\right\}$
Using the linearity property, we can split the Laplace transform of the sum into the sum of the Laplace transforms:
$ℒ\left\{\frac{1}{2}\right\}+ℒ\left\{\frac{1}{2}\mathrm{cos}\left(2x\right)\right\}$
Now, we can evaluate each term separately.
The Laplace transform of a constant function $c$ is given by:
$ℒ\left\{c\right\}=\frac{c}{s}$
Applying this property, we have:
$ℒ\left\{\frac{1}{2}\right\}=\frac{1}{2s}$
Next, we use the property of the Laplace transform for $\mathrm{cos}\left(2x\right)$.
The Laplace transform of $\mathrm{cos}\left(ax\right)$ is given by:
$ℒ\left\{\mathrm{cos}\left(ax\right)\right\}=\frac{s}{{s}^{2}+{a}^{2}}$
In this case, we have $\mathrm{cos}\left(2x\right)$, so $a=2$. Substituting the values, we get:
$ℒ\left\{\frac{1}{2}\mathrm{cos}\left(2x\right)\right\}=\frac{s}{{s}^{2}+{2}^{2}}=\frac{s}{{s}^{2}+4}$
Finally, we can combine the results:
$ℒ\left\{{\mathrm{cos}}^{2}\left(x\right)\right\}=\frac{1}{2s}+\frac{s}{{s}^{2}+4}$
Therefore, the Laplace transform of ${\mathrm{cos}}^{2}\left(x\right)$ is $\frac{1}{2s}+\frac{s}{{s}^{2}+4}$.

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