GATE | GATE CS 2011 | Question 31

Given i=√-1, what will be the evaluation of the integral \int_{0}^{\pi/2} \frac{\cos x + i\sin x}{\cos x - i\sin x} dx ?
(A) 0
(B) 2
(C) -i
(D) i


Answer: (D)

Explanation: 

  \int_{0}^{\pi/2} \frac{\cos x + i\sin x}{\cos x - i\sin x} dx \newline \newline = \int_{0}^{\pi/2} \frac{{e}^{ix}}{{e}^{-ix}}  dx  \hspace{0.2cm}= \int_{0}^{\pi/2} {e}^{2ix} dx \newline \newline = \frac{1}{2i} [{e}^{2ix}]^{\pi/2}_{0} \hspace{0.2cm} = \frac{1}{2i} [{e}^{i\pi} - {e}^{0}]  \newline \newline =  \frac{1}{2i} [\cos \pi + i\sin \pi - 1] \newline \newline = \frac{1}{2i}[-1 + 0 - 1] \times  \frac{i}{i} = i


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