IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

K4FECN6

You will have to use complex numbers to solve this integral.

The integral can be represented as the real part of e^cos&{cos(sin&)+isin(sin&)} d&

=real part of e^cos&.e^isin&d&

=real part of e^cos&+isin& d&

=real part of e power e^i& d&

=real part of {1+ e^i&/1! + e^2i&/2! + ...... } d&

={1+ cos&/1! + cos2&/2! + ......} d& [considering the real part of the above expr.]

=& + sin& + (1/2!)[sin2&/2] + ...... -----> 1

Now applying the proper limits we get the first term of 1 to be 2 pi and all other terms 0.

Cheers!!!!!!!

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