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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 | 382 views

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

selected

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*}

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?
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.