Question

l{sin^3 2t}

Answer 1

L=laplace?

Answer 2

i guess so

Answer 3

ok then first use
\[\sin 3A=3\sin!-4\sin^3A\]

Answer 4

then
\[\sin6A=3\sin2A-4\sin^32A\]

Answer 5

so
\[\sin^32t=(2\sin2t-\sin6t)/4\\so~L[\sin^{3}2t]=\frac{ L }{ 4 }[2\sin2t-\sin6t]\]

Answer 6

@UnkleRhaukus (◊���1=2=0�◊�)
is it ok?

Answer 7

there is no spoon

Answer 8

yeah lol

Answer 9

now we have to use
\[L[sinat]=\frac{ a }{ s^2 +a^2}\]