GATE CS 2011 Q.31 (CS Q.31): Calculus

Given $i = \sqrt{-1}$ , what will be the evaluation of the definite integral

\int_0^{\pi/2} \frac{\cos x + i \sin x}{\cos x - i \sin x} \, dx\text{ ?}

$-i$

$i$

Official answer and solution

Solution

Simplify the integrand using Euler's formula $e^{ix} = \cos x + i \sin x$ and integrate.

By Euler's formula, the numerator is $\cos x + i \sin x = e^{ix}$ and the denominator is $\cos x - i \sin x = e^{-ix}$ .

\frac{\cos x + i \sin x}{\cos x - i \sin x} = \frac{e^{ix}}{e^{-ix}} = e^{2ix}

\int_0^{\pi/2} e^{2ix} \, dx = \left[ \frac{e^{2ix}}{2i} \right]_0^{\pi/2} = \frac{e^{2i(\pi/2)} - e^0}{2i} = \frac{e^{i\pi} - 1}{2i}

\frac{-1 - 1}{2i} = \frac{-2}{2i} = -\frac{1}{i} = -(-i) = i

Final answer

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