Question

Givne $i=\sqrt{-1}$, what will be the evaluation of the definite intgral ${\int }_0^{\frac{\pi }{2}}\frac{cosx+isinx}{cosx-isinx}dx$

• A. $0$
• B. $2$
• C. $-i$
• D. $i$

Answer 1

The correct option is D $i$

We know that by Euler Notation,
$e^{i\theta }=cos\theta +isin\theta$ & $e^{-i\theta }=cos\theta -isin\theta$
So, $I={\int }_0^{\frac{\pi }{2}}\frac{cosx+isinx}{cosx-isinx}dx={\int }_0^{\frac{\pi }{2}}\frac{e^{ix}}{e^{-ix}}dx$
$={\int }_0^{\pi /2}{e}^{2ix}dx={\left[\frac{e^{2ix}}{2i}\right]}_o^{\pi /2}$
$=\frac{1}{2i}[e^{-\pi i}-1]=\frac{1}{2i}[-1-1]=\frac{-1}{i}$
$=\frac{-i}{i^2}=i$