Question

Given, \(i = \sqrt { - 1} \), the value of the definite integral, \(I = \mathop \smallint \limits_0^{\pi /2} \frac{{\cos x + i\sin x}}{{\cos x - i\sin x}}dx\) is:

1. 1
2. -1
3. i
4. -i

Answer 1

Correct Answer - Option 3 : i

Concept:

Some important relations

1) \({{\rm{e}}^{{\rm{ix}}}} = \cos {\rm{x}} + {\rm{i}}\sin {\rm{x\;\;}}\)

2) \({{\rm{e}}^{ - {\rm{ix}}}} = \cos {\rm{x}} - {\rm{i}}\sin {\rm{x}}\)

3) \({{\rm{e}}^{{\rm{i\pi }}}} = \cos {\rm{\pi }} + {\rm{i}}\sin {\rm{\pi }} = - 1\)

4) \(\frac{{{\rm{ - 1}}}}{{\rm{i}}}{\rm{ = \;i}}\)

Calculation:

\(\begin{array}{l} I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{\rm{cosx \ +\ isinx}}}}{{{\rm{cosx \ -\ isinx}}\;}}{\rm{dx}}\\ = \mathop \smallint \limits_0^{\frac{\pi }{0}} \frac{{{e^{ix}}}}{{{e^{ - ix}}}}dx =\mathop \smallint \limits_0^{\frac{\pi }{2}} {e^{2ix}}dx\\ = \frac{1}{{2i}}\left[ {{e^{2ix}}} \right]_0^{\frac{\pi }{2}} = \frac{1}{{2i}}\left[ {{e^{i\pi }} - 1} \right] \\= \frac{1}{{2i}}\left( { - 1 - 1} \right)\\ = - \frac{1}{i} \\= i \end{array}\)