We know that
$$ an t=frac{e^{it}-e^{-it}}{i(e^{it}+e^{-it})}=frac{e^{2i
t}+1-2}{i(e^{2it}+1)}=-i(1-frac{2}{e^{2it}+1}).$$
Now we try to find the anti-derivative of
$$
f(t)=frac{1}{e^{2it}+1}.
$$
We observe that
$$
frac{1}{e^{2it}+1}=frac{1}{(e^{it}+i)(e^{it}-i)}=-2i(frac{1}{e^{it}+i}-frac{1}{e^{it}-i}).
$$
So we try to find the anti-derivative of
$$
frac{1}{e^{it}+i},frac{1}{e^{it}-i}.
$$
This is hard,so we give up this way.