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

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

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

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

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

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

$=i$
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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?
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