GATE CSE 2011 | Question: 31

Question

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

• A. $0$
• B. $2$
• C. $-i$
• D. $i$

Comments

f(X)

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• $e^{i\theta} = \cos\theta + i \sin\theta$
• $e^{-i\theta} = \cos\theta - i \sin\theta$

Reference:

• https://math.stackexchange.com/questions/2735303/does-eulers-formula-give-e-ix-cosx-i-sinx 

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cos 180 = cos(90 + 90) = - sin 90 = -1

School maths

90, 270 => Change(sin <=> cos, tan <=> cot, sec <=> cosec)

Sign, 

1st Quadrant(90-x,) = all positive

2nd(90+x, 180-x) = sin, cosec

3rd (180+x, 270-x)= tan, cot

4th (270+x, 360-x) = cos, sec

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.

Answer 1

Answer is D.

$\int_{0}^{\frac{\pi}{2}}\frac{e^{ix}}{e^{-ix}}dx \\= \int_{0}^{\frac{\pi}{2}}e^{2ix}dx \\= \dfrac{e^{2ix}}{2i}\mid_{0}^{\frac{\pi}{2}} \\= \dfrac{-2}{2i} = \dfrac{-1}{i} = \dfrac{-1\times i}{i\times i}=\dfrac{-i}{i^2}=\dfrac{-i}{-1}=i$

Comments

To rationalize, multiply numerator and denominator by $cosx+isinx$ in this case.

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Note: e^(iπ) = -1

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NOTE:  $\cos(2x) + i\sin(2x) = e^{i2x}$

Answer 2

we know that 

also, 

Edit is neccessary$:$ Third last step $sin(\pi)$ instead of $sinx$

answer = option D

Comments

how is  -2/2i = i ?

where did - ve sign go ?

Please explain.

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@Jhaiyam

$\dfrac{-2}{2i} = \dfrac{-2i}{2i^{2}} = \dfrac{-2i}{2(-1)} = i\:\:\:[\because i = \sqrt{-1}, i^{2} = -1]$

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Thanks.

Answer 3

answer is i.

Comments

THANKS DEAR

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Best answer, i probably won't remember the other formula.

Answer 4

e^ix=cos x + i  sin x

e^-ix=cos x - i sin x

given integral can be written as I= ∫e^ix/e^-ix  dx   =∫e^2ix dx

I=e^2ix/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

Comments

complex number is in our syllabus ?

Answer 5

$\int_{0}^{\frac{\pi }{2}}\frac{\left ( cosx+isinx \right )^{2}}{cos^{2}x+sin^{2}x}dx$

$=\int_{0}^{\frac{\pi }{2}}\left ( cos^{2}x-sin^{2}x+isin2x \right )dx$

$=\int_{0}^{\frac{\pi }{2}}\left ( cos2x+isin2x \right )dx$

$=\left [ \frac{sin2x}{2}-\frac{icos2x}{2} \right ]_{0}^{\frac{\pi }{2}}$

$= \frac{\sin \pi }{2}-i\frac{\cos \pi }{2}-\left ( 0-\frac{i}{2} \right )$

$=-\left ( \frac{-i}{2} \right )+\frac{i}{2}$

$=i$

Comments

In the 2nd line it should be :

$cos^2x - sin^2x+isin2x$

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@srestha ma'am

please edit the answer.

$(cosx+isinx)^{2}=cos^{2}x+i^{2}sin^{2}x+2icosx.sinx$

$(cosx+isinx)^{2}=cos^{2}x-sin^{2}x+2isinx.cosx$              Because $i=\sqrt{-1},i^{2}=-1$

$(cosx+isinx)^{2}=cos2x+2isinx.cosx$                               Because $cos2x=cos^{2}x-sin^{2}x$

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ok, now?

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Yes correct!

Answer 6

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?

why?

Comments

2sin x cos x = sin 2x, not cos 2x..

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oh sorry. yes sin2x.

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@ 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.