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+2 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 (73.4k points)   | 302 views

3 Answers

+9 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 (4k points)  
selected by
+5 votes

we know that 


answer = option D

answered by Veteran (26.9k 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 (2.7k 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.
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