# Laplace transform question

1. Oct 2, 2007

### engineer_dave

1. The problem statement, all variables and given/known data

Find the Laplace transform of the following functions

f(t)= sine squared t

2. Relevant equations

sin kt = k^2/(s^2 + k^2)

3. The attempt at a solution

I went like this;

sin squared t is sin t* sin t therefore it should be 1/(s^2 +1) (s^2 + 1) but unfortunately this was not the answer! any help would be very much appreciated. Thanks

2. Oct 2, 2007

### cristo

Staff Emeritus
Well, the Laplace transform of a function is defined by an integral, isn't it? Could you write this integral, and try and solve it?

3. Oct 2, 2007

### engineer_dave

what do u mean?

4. Oct 2, 2007

### engineer_dave

u mean it would be the integral of e^-st multiplied by sine squared t??

5. Oct 2, 2007

### cristo

Staff Emeritus
Yes. Either that, or there may be a useful theorem you can use. I can't, however, remember whether the theorem I'm thinking of concerns the Laplace transform of products, or the product of Laplace transforms.

6. Oct 2, 2007

### real10

Hint use the trig indentity
$$sin^2(x)=\frac{1-cos(2x)}{2}$$
using this u get $$\frac{1}{2}*({\frac{1}{s}-\frac{s}{s^2+4})$$

which simplifies to the answer -> $$\frac{2}{s^3+4s}$$

7. May 29, 2010

### MorrisonHotel

I think it was the theorem of the convolution operator, which is to be found at http://en.wikipedia.org/wiki/Convolution" [Broken]

@engineer_dave; remember L{ f.g } is not equal to L{f} . L{g} ...However the convolution operator makes it so. Read the stuff at the link. Or the quickest way (since you're an engineer, it's the one you might be interested in) as real10 mentioned;

sin^2(t) = 1/2 - cos(2t)/2
=> L{sin^2(t)} = 1/2. L{1} - 1/2. L{cos2t} (=> due to the linearity of laplace operation, easily provable)
= 1/2 . 1/s - 1/2. 1/s^2 + 4
= 4 / s^3 + 4s indeed..

Last edited by a moderator: May 4, 2017
8. Jun 2, 2010