Given i = √(-1), what will be the evaluation of the definite integral 0 /2 (cos x + i sin x)/(cos x - i sin x) dx? A) 0 B) 2 C) -i D) i

Official answer: D) i. Explanation: Using Euler's formula: cos x + i sin x = eix cos x - i sin x = e-ix The integrand simplifies to: (cos x + i sin x)/(cos x - i sin x) = eixe-ix = e2ix Now, evaluate the definite integral: I = 0 /2 e2ix dx = [ e2ix2i ]0 /2 Substituting the upper and lower limits: I = e2i( /2) - e02i = ei - 12i Since ei = cos + i sin = -1: I = (-1 - 1)/(2i) = (-2)/(2i) = -(1)/(i) = -(-i) = i Thus, the correct option is (D) [p1].

Calculus

GATE CSE 2011 · Q31

MCQ2 marksEasy

Time00:00

Given $i = \sqrt{-1}$ , what will be the evaluation of the definite integral $\int_{0}^{\pi/2} \frac{\cos x + i \sin x}{\cos x - i \sin x}\, dx$ ?

Select one answerSingle correct

Pick an option to check

Solution

Step-by-step verified solution

Official answer: D) i

Calculus

Using Euler's formula:

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

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

The integrand simplifies to:

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

Now, evaluate the definite integral:

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

Since $e^{i\pi} = \cos \pi + i \sin \pi = -1$ :

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

Thus, the correct option is (D) [p1].

Try the question first, then check your work.

Think the official key or this write-up is wrong?

Report error or challenge

Examiner trap

The misconception this question tests

This Calculus previous year question is Q31 of the GATE CSE 2011 paper. Find more GATE CSE PYQs on Calculus in Engineering Mathematics, or browse all GATE CSE PYQs.

Free diagnostic

Where do you stand for GATE?

Take the 20-minute diagnostic: the 3 topics that cost you the most marks, and a rank range. Or sit 5 related Calculus PYQs now with no account.

One GATE CSE PYQ every morning

An official question by email at 7 AM IST, with a link to its verified solution. Unsubscribe in one click.

Calculus in GATE CSE

Calculus has 40 GATE CSE PYQs across 25 papers from 2011 to 2026. 8 of them are 2-mark MCQs like this one, and it is the only one in GATE CSE 2011.

How GATE tests Calculus

More from GATE CSE 2011

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.