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Integral Calculus

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8 Feb 2007 14:36:25 IST

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A CHALLENGING ONE!!!

None

Evaluate

(e^ix-e^-ix)dx/(i(e^ix+e^-ix))

Its really simple

So solve quickly

Cheers

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Comments (11)

AAKRITI

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8 Feb 2007 14:54:47 IST

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1/2 [ ln {mod( e^2ix/(1+e^2ix) ) } - ln(e^2ix+1) ]

Anshuman Johri

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8 Feb 2007 15:13:46 IST

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Is the answer 1?

CyBorG

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8 Feb 2007 15:15:28 IST

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No.The ans is not 1.Its really simple.And plz see that I'm not expecting ans in terms of complex nos.

AAKRITI

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8 Feb 2007 15:23:06 IST

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-log[mod(e^ix+e^-ix)] + c , i am sure

vinu

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8 Feb 2007 15:33:13 IST

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Hi kab,

The ans is.......log(sec x) .

himanshuagrawal

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8 Feb 2007 15:39:31 IST

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The ans is

-log(e^ix+e^-ix)+c

himanshuagrawal

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8 Feb 2007 15:41:27 IST

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ans. is

-log(2cosix)+c

CyBorG

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8 Feb 2007 16:03:10 IST

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Vinu has stated the perfect answer.The other answers may be correct.

The easiest way to solve is this

We know

sinx=(e^ix-e^-ix)/(2i)

and cosx =(e^ix+e^-ix)/(2)

So given question is nothing but

tanxdx=log(secx)

Cheers!!!!!

AAKRITI

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8 Feb 2007 16:07:27 IST

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hi adarsh , one answer can be always expressed in many forms

mine is also correct

CyBorG

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8 Feb 2007 16:12:19 IST

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That is what I meant by saying "the other ans may be correct".
But the shortest is that given by me and vinu!!!

AAKRITI

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8 Feb 2007 16:25:16 IST

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