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NDA 2025
Exam
NDA
Year
2025
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2025 · Mathematics · Indefinite Integration

Consider the following for the two (02) items that follow: Let the...

Consider the following for the two (02) items that follow:
Let the function y=(1−cos⁡x)−1y = (1 - \cos x)^{-1}, where x≠2nπx \neq 2n\pi and nn is an integer.

What is ∫y dx\int y \, dx equal to?

Options

  1. A−tan⁡(x/2)+c-\tan(x/2) + c
  2. B−cot⁡(x/2)+c-\cot(x/2) + c
  3. Ctan⁡(x/2)+c\tan(x/2) + c
  4. Dcot⁡(x/2)+c\cot(x/2) + c

Correct answer

B. −cot⁡(x/2)+c-\cot(x/2) + c

Step-by-step solution

Given y=(1−cos⁡x)−1y = (1 - \cos x)^{-1}

y=11−cos⁡xy = \dfrac{1}{1 - \cos x}

Using the half-angle formula 1−cos⁡x=2sin⁡2(x/2)1 - \cos x = 2 \sin^2(x/2)

y=12sin⁡2(x/2)=12csc⁡2(x/2)y = \dfrac{1}{2 \sin^2(x/2)} = \dfrac{1}{2} \csc^2(x/2)

Integrating with respect to xx:

∫y dx=∫12csc⁡2(x/2) dx\int y \, dx = \int \dfrac{1}{2} \csc^2(x/2) \, dx

∫y dx=12(−cot⁡(x/2)1/2)+c\int y \, dx = \dfrac{1}{2} \left( \dfrac{-\cot(x/2)}{1/2} \right) + c

∫y dx=−cot⁡(x/2)+c\int y \, dx = -\cot(x/2) + c

Answer: −cot⁡(x/2)+c-\cot(x/2) + c
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