What is the Laplace transform of sin2tcos2t?

[RafiS]

iv got a problem i can't seem to understand. if anyone could help me out it would be great

f(t)= sin2tcos2t

im just not sure what to do when i have the product of 2 trig functions.

the correct answer is (2/(s^2 + 16))

thanks for any help

 

[exk]

Is there something you can do with f(t) to make it look different?

 

[sutupidmath]

yeah, like exk suggested, try to use this trig identity somehow:

sin2x=2isnxcosx can you figure it out how to transform your f(t) into a similar form?

 

[Vid]

Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)

 

[sutupidmath]

Use f(x) = 1/4sin^2(2t)
f'(x) = sin(2t)cos(2t)

I don't see how would this help!
If i have gotten the op right, he just needst to take the laplace transform of

f(t)=sin2tcos2t=\frac{1}{2}sin4t
so

L{f(t)}=L{\frac{1}{2}sin4t}=\int_{0}^{\infty}\frac{1}{2}sin(4t)e^{-st}dt

 

[Vid]

I was thinking that sin^2(t) was a common laplace transform, but I was mistaken.

 

[exk]

another strategy for this would be to look up in a transform table what your answer corresponds to and work backwards.

sutupidmath's solution is correct.