Lcos2x

Question

Lcos2x

xleb123 · Accepted Answer

To solve the given problem, which is finding the Laplace transform of

\cos^{2} (x)

, we'll apply the definition and properties of the Laplace transform.
The Laplace transform of a function

f (t)

is defined as:

ℒ {f (t)} = F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t

In this case, we want to find the Laplace transform of

\cos^{2} (x)

. However, the Laplace transform is typically used for functions defined on the interval

[0, \infty)

. The function

\cos^{2} (x)

is defined for all real values of

x

. To proceed, we need to express

\cos^{2} (x)

in terms of a function defined on the interval

[0, \infty)

.
Using the identity

\cos^{2} (x) = \frac{1}{2} (1 + \cos (2 x))

, we can rewrite

\cos^{2} (x)

as

\frac{1}{2} (1 + \cos (2 x))

.
Now, we can find the Laplace transform of

\cos^{2} (x)

by applying the linearity property of the Laplace transform.

ℒ {\cos^{2} (x)} = ℒ {\frac{1}{2} (1 + \cos (2 x))}

Using the linearity property, we can split the Laplace transform of the sum into the sum of the Laplace transforms:

ℒ {\frac{1}{2}} + ℒ {\frac{1}{2} \cos (2 x)}

Now, we can evaluate each term separately.
The Laplace transform of a constant function

c

is given by:

ℒ {c} = \frac{c}{s}

Applying this property, we have:

ℒ {\frac{1}{2}} = \frac{1}{2 s}

Next, we use the property of the Laplace transform for

\cos (2 x)

.
The Laplace transform of

\cos (a x)

is given by:

ℒ {\cos (a x)} = \frac{s}{s^{2} + a^{2}}

In this case, we have

\cos (2 x)

, so

a = 2

. Substituting the values, we get:

ℒ {\frac{1}{2} \cos (2 x)} = \frac{s}{s^{2} + 2^{2}} = \frac{s}{s^{2} + 4}

Finally, we can combine the results:

ℒ {\cos^{2} (x)} = \frac{1}{2 s} + \frac{s}{s^{2} + 4}

Therefore, the Laplace transform of

\cos^{2} (x)

is

\frac{1}{2 s} + \frac{s}{s^{2} + 4}

.

Answered question