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Given $i = \sqrt{-1}$, what will be the evaluation of the definite integral $\int \limits_0^{\pi/2} \dfrac{\cos x +i \sin x} {\cos x - i \sin x} dx$ ?

  1. $0$
  2. $2$
  3. $-i$
  4. $i$
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Best answer
45 votes
45 votes

Answer is D.

$\int_{0}^{\frac{\pi}{2}}\frac{e^{ix}}{e^{-ix}}dx \\= \int_{0}^{\frac{\pi}{2}}e^{2ix}dx \\= \dfrac{e^{2ix}}{2i}\mid_{0}^{\frac{\pi}{2}} \\= \dfrac{-2}{2i} = \dfrac{-1}{i} = \dfrac{-1\times i}{i\times i}=\dfrac{-i}{i^2}=\dfrac{-i}{-1}=i$

36 votes
36 votes

we know that 

also, 

Edit is neccessary$:$ Third last step $sin(\pi)$ instead of $sinx$

answer = option D

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1 votes
1 votes

eix=cos x + i  sin x

e-ix=cos x - i sin x

given integral can be written as I= ∫eix/e-ix  dx   =∫e2ix dx

I=e2ix/2i

putting value 0 and π/2

we get e^π-e^0/2 i=(cos π+sin π-cos 0-i sin 0)/2i

                             =-2/2i

                             =-1/i

                             =i

Answer:

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