The best method to solve this is using complex integration.(I remember that I had solved this sum once).Let us take C=e^coskcos(sink)dk and S=e^cosksin(sink)dk

Now,C+iS=e^cosk[cos(sink)+isin(sink)]dk=e^cosk+isinkdk[since cosk+isink=e^ik]

Writing the expansion of e^x,We get,

[1+cisk/1!+(cisk)²/2!+(cisk)³/3!+.......]dk,Here (cisk)ⁿ=cisnk

C is the real part of this integral,So,

C=[1+cosk/1!+cos2k/2!+.....]dk

=(2pi-0)+(sin2pi-sin0)/1!+.......=2pi

Except the first term,All the remaining terms become 0.Hence the ans is 2