25 votes 25 votes Given $i = \sqrt{-1}$, what will be the evaluation of the definite integral $\int \limits_0^{\pi/2} \dfrac{\cos x +i \sin x} {\cos x - i \sin x} dx$ ? $0$ $2$ $-i$ $i$ Calculus gatecse-2011 calculus integration normal + – go_editor asked Sep 29, 2014 • edited Nov 27, 2017 by pavan singh go_editor 11.1k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Dileep kumar M 6 commented Nov 27, 2017 reply Follow Share f(X) 9 votes 9 votes Lakshman Bhaiya commented Nov 22, 2019 reply Follow Share $e^{i\theta} = \cos\theta + i \sin\theta$ $e^{-i\theta} = \cos\theta - i \sin\theta$ Reference: https://math.stackexchange.com/questions/2735303/does-eulers-formula-give-e-ix-cosx-i-sinx 6 votes 6 votes Please log in or register to add a comment.
Best answer 46 votes 46 votes Answer is D. $\int_{0}^{\frac{\pi}{2}}\frac{e^{ix}}{e^{-ix}}dx \\= \int_{0}^{\frac{\pi}{2}}e^{2ix}dx \\= \dfrac{e^{2ix}}{2i}\mid_{0}^{\frac{\pi}{2}} \\= \dfrac{-2}{2i} = \dfrac{-1}{i} = \dfrac{-1\times i}{i\times i}=\dfrac{-i}{i^2}=\dfrac{-i}{-1}=i$ sonapraneeth_a answered Jan 19, 2015 • edited Nov 14, 2019 by KUSHAGRA गुप्ता sonapraneeth_a comment Share Follow See all 4 Comments See all 4 4 Comments reply Ayush Upadhyaya commented Nov 26, 2017 reply Follow Share $\int_{0}^{\frac{\pi}{2}} \frac{\cos x+i\sin x}{cos x -isin x} dx$ After rationalising we get $\int_{0}^{\frac{\pi}{2}} cos^{2} x -sin^{2} x + i.2.sinxcosx dx$ Using identities $cos^{2}x -sin^{2} x= cos2x$ 2sinxcosx= sin2x integral becomes $\int_{0}^{\frac{\pi}{2}} cos 2x + i.sin 2x$ which on evaluating gives answer as i (option d) 25 votes 25 votes mehul vaidya commented Apr 22, 2018 reply Follow Share 0 . numerator is 0 and not -2. 0 votes 0 votes Jhaiyam commented Jul 29, 2020 reply Follow Share Lame question, but can you tell me how to rationalize the denominator in this case ? 0 votes 0 votes Gupta731 commented Sep 29, 2020 reply Follow Share To rationalize, multiply numerator and denominator by $cosx+isinx$ in this case. 0 votes 0 votes Please log in or register to add a comment.
36 votes 36 votes we know that also, Edit is neccessary$:$ Third last step $sin(\pi)$ instead of $sinx$ answer = option D amarVashishth answered Oct 22, 2015 • edited Jan 26, 2019 by Lakshman Bhaiya amarVashishth comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Jhaiyam commented Apr 22, 2020 reply Follow Share how is -2/2i = i ? where did - ve sign go ? Please explain. 0 votes 0 votes Lakshman Bhaiya commented Apr 23, 2020 reply Follow Share @Jhaiyam $\dfrac{-2}{2i} = \dfrac{-2i}{2i^{2}} = \dfrac{-2i}{2(-1)} = i\:\:\:[\because i = \sqrt{-1}, i^{2} = -1]$ 2 votes 2 votes Jhaiyam commented Apr 28, 2020 reply Follow Share Thanks. 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes answer is i. resuscitate answered Nov 4, 2015 resuscitate comment Share Follow See all 2 Comments See all 2 2 Comments reply monali commented Nov 4, 2015 reply Follow Share THANKS DEAR 0 votes 0 votes LiteYagami commented Jan 19 reply Follow Share Best answer, i probably won't remember the other formula. 0 votes 0 votes Please log in or register to add a comment.
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 Pooja Palod answered Nov 4, 2015 Pooja Palod comment Share Follow See 1 comment See all 1 1 comment reply tiger commented Nov 28, 2015 reply Follow Share complex number is in our syllabus ? 0 votes 0 votes Please log in or register to add a comment.