See attached for questions
a) L [ Sin t / t ]
L[ Sin t ] = 1/(s^2+1)
Now, if f(t) <-----> F(s)
then, f(t)/t <-----> Integral (from s to Infinity) [F(s)ds]
So, L [ Sin t / t ] = Integral (from s to Infinity) [ds/(s^2+1)]
L [ Sin t / t ] = tan^-1 (1/s)...( "tan ...
There area variety of Laplace transformations in this solution