Skip to content
Related Articles

Related Articles

Improve Article

GATE | GATE CS 2011 | Question 31

  • Difficulty Level : Easy
  • Last Updated : 14 Feb, 2018

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     


Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :