I want to take the inverse laplace transform of $$\frac{e^{-s}}{s(s^{2}+1)}$$
So I separate the equation into $$e^{-s}\times\frac{1}{s(s^{2}+1)}$$
Now, I take the partial fraction of $\frac{1}{s(s^{2}+1)}$.
I get $$\frac{A}{s} +\frac{Bs+C}{s^{2}+1}$$
$$\frac{A(s^{2}+1)+(Bs+C)s}{s(s^{2}+1)}$$
After solving for A,B,C I get $A=1$,$B=-1$,$C=0$
And substituting back to the equation, I get $$\frac{1}{s} -\frac{1}{s^{2}+1}$$
Then by the 2nd shifting property,$$\mathcal{L}(e^{-s}\times\frac{1}{s} -\frac{1}{s^{2}+1})$$
My final answer is $$u(t-1)(1-sin(t-1))$$ but the answer provided is $$u(t-1)(1-cos(t-1))$$
Could someone help me verify? Thanks