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

1. $0$
2. $2$
3. $-i$
4. $i$
edited | 2k views
+2

f(X)

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

selected
+8
$\int_{0}^{\frac{\pi}{2}} \frac{\cos x+i\sin x}{cos x -isin x} dx$

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)
0
0 . numerator is 0 and not -2.

we know that \large \begin{align*} \cos{x} + i\sin{x} &= e^{ix} \\ \end{align*}

also,

\begin{align*} \int \limits_0^{\pi/2} \frac{\cos x +i \sin x} {\cos x - i \sin x}\ dx&= \int_{0}^{\pi /2}\frac{e^{ix}}{e^{-ix}}\ dx \\ &= \int_{0}^{\pi /2}e^{2ix}\ dx\\ &= \left[ \frac{e^{2ix}}{2i} \right ]_{0}^{\pi /2}\\ &= \frac{1}{2i} \left[ e^{2i \times (\pi /2)} -1 \right ] \\ &= \frac{\cos \pi + i\sin x - 1}{2i}\\ &= \frac{-2}{2i}\\ &= i \end{align*}

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

edited
+1 vote
$\int_{0}^{\frac{\pi }{2}}\frac{\left ( cosx+isinx \right )^{2}}{cos^{2}x+sin^{2}x}dx$

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

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

$=\frac{sin2x}{2}-\frac{cos2x}{2}$

$=\frac{i}{2}-\frac{-i}{2}$

$=i$
+2
In the 2nd line it should be :

$cos^2x - sin^2x+isin2x$
0

@ ma'am

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

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

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

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?
0
2sin x cos x = sin 2x, not cos 2x..
+1
oh sorry. yes sin2x.
0
@ 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.
–1 vote
file:///C:/Users/ram/Desktop/20180728_151152.jpg
+2

OK

+1 vote
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