ME , GATE2006

Question

Assuming $i=\sqrt{-1}$ and t is real number , $\int_{0}^{\Pi /3} e^{it}dt$

given ans is $\sqrt{3}/2 + i/2 , and i am getting \sqrt{3}/2-i/2$

i solved first put $e^{it}= cost +isint$ then and integrate , but in solution they first integrate $e^{it}$ then put the value of it ... but ans different ... why is this happening .. am i missing something ??

Comments

∫e^itdt on limit  [Π/3,0]
= ∫cost+isint dt
=[sint - icost]
put limit (sqrt(3)/2 -i/2) -(0-i)=sqrt(3)/2+i/2

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ok thanks got my mistake

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Hi @sid1221 ji,

Please mention answer and close the question if ur doubt is resolved.