Question

In Problems 37–40 find $L\left\{f\left(t\right)\right\}$by first using a trigonometric identity.

38.$f\left(t\right)={\cos }^{2}t$

Short answer

The Laplace transform of given function is,

$L\left\{{\cos }^{2}t\right\}=\frac{1}{2s}+\frac{s}{2\left(s^2+4\right)},s>0$

Step-by-step solution

Theorem

Here we determine the Laplace transform of the given function by comparing it to general form as provided in theorem 7.1.1.

$L\left\{f\left(t^n\right)\right\}=\frac{n!}{s^{n+1}},s>0$

$L\left\{\cos at\right\}=\frac{s}{s^2+a^2},s>a$

Solution

Consider the function$f\left(t\right)={\cos }^{2}t$

The objective is to find $L\left\{f\left(t\right)\right\}$using the theorem.

From the linearity property of the Laplace transform, we have

$$\begin{gathered} f\left(t\right)=\cos {}^{2}t \\ f\left(t\right)=\frac{1}{2}\left(1+\cos 2t\right)............................\left(\because \cos 2\theta =2{\cos }^2\theta -1\right) \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{1+\cos 2t\right\} \\ L\left\{f\left(t\right)\right\}=\frac{1}{2}L\left\{1\right\}+\frac{1}{2}L\left\{\cos 2t\right\} \end{gathered}$$

From the theorem7.1.1,

$L\left\{f\left(t^n\right)\right\}=\frac{n!}{s^{n+1}},s>0,L\left\{\cos at\right\}=\frac{s}{s^2+a^2},s>a$, Where$a=0,1,2,3,....,n=0,1,2,3,........,$

$$\begin{gathered} L\left\{{\cos }^{2}t\right\}=\frac{1}{2}.\frac{1}{s}+\frac{1}{2}\left(\frac{s}{s^2+2^2}\right) \\ L\left\{{\cos }^{2}t\right\}=\frac{1}{2s}+\frac{s}{2\left(s^2+4\right)},s>0 \end{gathered}$$

Therefore the required Laplace transform of function is,

$L\left\{{\cos }^{2}t\right\}=\frac{1}{2s}+\frac{s}{2\left(s^2+4\right)},s>0$