GATE CS 2011 Q31 — Maxima and Minima, Mean Value Theorem, Integration

We can simplify the integrand using Euler's formula, which states that:

$$\cos x + i \sin x = e^{ix}$$

$$\cos x - i \sin x = e^{-ix}$$

Substituting these into the given integral, we get:

$$I = \int_{0}^{\pi/2} \frac{e^{ix}}{e^{-ix}} \, dx = \int_{0}^{\pi/2} e^{2ix} \, dx$$

Now, we integrate with respect to $x$:

$$I = \left[ \frac{e^{2ix}}{2i} \right]_{0}^{\pi/2}$$

Substituting the upper and lower limits:

$$I = \frac{e^{2i(\pi/2)} - e^{0}}{2i} = \frac{e^{i\pi} - 1}{2i}$$

Using Euler's identity, $e^{i\pi} = \cos \pi + i \sin \pi = -1$. Substituting this value:

$$I = \frac{-1 - 1}{2i} = \frac{-2}{2i} = -\frac{1}{i}$$

Since $\frac{1}{i} = -i$, we have:

$$I = -(-i) = i$$

Therefore, the correct option is D.