Question

L[sin^4 (t)]

Answer 1

try to expand sin^4x first

Answer 2

\[\sin^4x=\frac{ (1-\cos2x)^2 }{ 4 }=\frac{ 1-2\cos2x+\cos^2(2x) }{ 4 }\]

Answer 3

now multiply 2 with Dr and Nr

Answer 4

\[\sin^4t=\frac{ 2-4\cos2t+2\cos^22t }{ 8}=\frac{ 2-4\cos2t+1+\cos4t }{ 8 }\\=\frac{ 3+\cos4t-4\cos(2t) }{ 8 }\\so\\L[\sin^4t]=L[3/8]+L[\frac{ \cos4t }{ 8 }]-L[\frac{ \cos2t }{ 2 }]\\=\frac{ 3}{ 8}s+\frac{ 1 }{ 8 }\frac{ s }{ s^2+16 }-\frac{ 1 }{ 2 }\frac{ s }{ s^2+4 }\]
now u can simplify a bit more :)
@deobaran

Answer 5

could always use some convolution integrals as well :)