Krdytkyu

I am given a following signal and want to calculate it's Fourier series coefficients:

$$ x(t) = left{
begin{array}{ll}
0 & quad 0< t < 2 \
2 & quad 2< t < 3 \
0& quad 3< t < 4 \
-2 & quad 5< t < 6
end{array}
right. $$

Following is my method:

$$ a_k= frac{1}{6} ( int_{2}^{3} 2 e^{-jkw_0t}dt + int_{5}^{6} (-2) e^{-jkw_0t}dt) $$

$$ a_k= frac{2}{6} ( int_{2}^{3} e^{-jkw_0t}dt - int_{5}^{6} e^{-jkw_0t}dt) $$

After doing some simplification we get:

$$ a_k= frac{1}{jkpi}(e^{-2jfrac{kpi}{3}}-e^{-jkpi}+e^{-2jkpi}-e^{-5jfrac{kpi}{3}}) $$

I am stuck after this last step.

Hint:
for these two terms
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}$$
let
$$alpha-x=-2jfrac{kpi}{3}~~~;~~~alpha+x=-5jfrac{kpi}{3}$$
then
$$alpha=-7jfrac{kpi}{6}~~~;~~~x=-jfrac{kpi}{2}$$
and therefore
$$e^{-2jfrac{kpi}{3}}-e^{-5jfrac{kpi}{3}}=color{blue}{2je^{-7jkpi/6}sinfrac{kpi}{2}}$$