Question

Section 8.1 Introduction to the Laplace Transforms

Problem 2.
Use the table of Laplace transforms to find the Laplace transforms of the following functions.
$(a)cosh \: t \: sin \: t$
$(b) {sin}^{2} t$
$(c) {cos}^{2} 2t$
$(d) {cosh}^{2} t$
$(e)t \: sinh \: 2t$
$(f)sin \: t \: cos \: t$
$(g)sin(t + \frac{\pi}{4} )$
$(h)cos2t - cos3t$
$(i)sin2t + cos4t$
​

Answer 1

I don't know what table you have as reference, but I suspect it includes the following transforms:

$1 \leftrightarrow \dfrac1s$

$e^{at} \leftrightarrow \dfrac{1}{s-a}$

$\cos(at) \leftrightarrow \dfrac{s}{s^2+a^2}$

$\sin(at) \leftrightarrow \dfrac{a}{s^2+a^2}$

$\cosh(at) \leftrigharrow \dfrac{s}{s^2-a^2}$

$\sinh(at) \leftrigharrow \dfrac{a}{s^2-a^2}$

It probably also includes some more general properties, like

$t f(t) \leftrightarrow -F'(s)$

$e^{at} f(t) \leftrightarrow F(s-a)$

where F(s) is the Laplace transform of f(t).

Beyond these, you should also know the following identities:

$\cosh(t) = \dfrac{e^t + e^{-t}}2$

$\cosh^2(t) = \dfrac{1 + \cosh(2t)}2$

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

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

$\sin(2t) = 2 \sin(t) \cos(t)$

$\sin(t \pm T) = \sin(t) \cos(T) \pm \cos(t) \sin(T)$

Putting everything together, we have

• (a)

$\cosh(t) \sin(t) = \dfrac{e^t + e^{-t}}2 \times \sin(t) = \dfrac12 e^t \sin(t) + \dfrac12 e^{-t} \sin(t)$

and the Laplace transform is

$\dfrac12 F(s - 1) + \dfrac12 F(s + 1)$

where F(s) is the transform of sin(t),

$F(s) = \dfrac{1}{s^2 + 1}$

Then

$\cosh(t) \sin(t) \leftrightarrow \dfrac{\frac1{(s-1)^2+1} + \frac1{(s+1)^2+1}}2 = \boxed{\dfrac{s^2+2}{s^4+4}}$

• (b)

$\sin^2(t) = \dfrac12 \left(1 - \cos(2t)\right)$

and the transform is

$F(s) = \dfrac12 \left(\dfrac1s - \dfrac{s}{s^2+4}\right) = \boxed{\dfrac{2}{s^3+4s}}$

• (c)

$\cos^2(2t) = \dfrac12 \left(1 + \cos(4t)\right)$

with transform

$F(s) = \dfrac12 \left(\dfrac1s + \dfrac{s}{s^2+16}\right) = \boxed{\dfrac{s^2+8}{s^3+16s}}$

• (d)

$\cosh^2(t) = \dfrac12 \left(1 + \cosh(2t)\right)$

with transform

$F(s) = \dfrac12 \left(\dfrac1s + \dfrac{s}{s^2-4}\right) = \boxed{-\dfrac2{s^3-4s}}$

• (e)

$t\sinh(2t) \leftrightarrow -F'(s)$

where F(s) is the Laplace transform of sinh(2t),

$F(s) = \dfrac{2}{s^2 - 4} \implies -F'(s) = \boxed{\dfrac{4s}{(s^2-4)^2}}$

• (f)

$\sin(t) \cos(t) = \dfrac12 \left(2\sin(t) \cos(t)\right) = \dfrac12 \sin(2t)$

with transform

$F(s) = \dfrac12 \times \dfrac{2}{s^2+4} = \boxed{\dfrac1{s^2+4}}$

• (g)

$\sin\left(t+\dfrac\pi4\right) = \sin(t) \cos\left(\dfrac\pi4\right) + \cos(t) \sin\left(\dfrac\pi4\right) = \dfrac1{\sqrt2} \left(\sin(t) + \cos(t)\right)$

with transform

$F(s) = \dfrac1{\sqrt2} \left(\dfrac1{s^2+1} + \dfrac{s}{s^2+1}\right) = \boxed{\dfrac{s+1}{\sqrt2 (s^2+1)}}$

The last two are trivial and follow directly from the properties listed above.

• (h)

$\cos(2t) - \cos(3t) \leftrightarrow \dfrac{s}{s^2+4} - \dfrac{s}{s^2+9} = \boxed{\dfrac{5s}{s^4 + 13s^2 + 36}}$

• (i)

$\sin(2t) + \cos(4t) \leftrightarrow \dfrac2{s^2+4} + \dfrac{s}{s^2+16} = \boxed{\dfrac{s^3+2s^2+4s+32}{s^4+20s^2+64}}$