First time here? Checkout the FAQ!
+3 votes

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

(A) 0

(B) 2

(C) $-i$

(D) $i$

asked in Calculus by Veteran (76k points)   | 390 views

3 Answers

+10 votes
Best answer

Answer is D.

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


answered by Loyal (3.9k points)  
selected by
+5 votes

we know that 


answer = option D

answered by Veteran (27.5k points)  
0 votes
Great answers down this one but..
Somebody please explain me why is this trivial approach wrong:
rationalise the fraction. denominator will become 1.

apply f(x)=f(a-x) formula of definite integration thereafter. and add the two integrals. the sqr. terms cancel out.

the final integration we get is:
0 to pi/2 integral( 2i sinx cosx)

this is 0 to pi/2 integral i.cos2x.

treating i as constant is wrong?

answered by Loyal (2.8k points)  
2sin x cos x = sin 2x, not cos 2x..
oh sorry. yes sin2x.
@ Aspi,

Can you please explain what is definite integration formula you mentioned above f(x)=f(a-x).

I am rationalizing numerator and denominator by cosx+isinx. After that i am getting

Integral 0 to pi/2   1 - 2sin^2x - isin2x. I don't know how to proceed from here.

Top Users Mar 2017
  1. rude

    4018 Points

  2. sh!va

    2994 Points

  3. Rahul Jain25

    2804 Points

  4. Kapil

    2608 Points

  5. Debashish Deka

    2104 Points

  6. 2018

    1414 Points

  7. Vignesh Sekar

    1336 Points

  8. Bikram

    1218 Points

  9. Akriti sood

    1186 Points

  10. Sanjay Sharma

    1016 Points

Monthly Topper: Rs. 500 gift card

21,446 questions
26,759 answers
22,955 users