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UPSC NDA I 2025 Mathematics Paper

Question Numbers: 81-82

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

Let the function

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

where

x \neq 2n\pi

and

n

is an integer

Question : 82 of 120

Marks: +1, -0

What is

∫ydx

c

is the constant of integration.

Solution:

Concept:

Use the trigonometric identity

1-\cos x = 2\sin^2\left(\frac{x}{2}\right)

to rewrite the integrand in terms of

\csc^2\left(\frac{x}{2}\right)

, then integrate using the standard result

\int \csc^2 u \, du = -\cot u + C

Explanation:

Given

y = \frac{1}{1-\cos x}

Using

1-\cos x = 2\sin^2\left(\frac{x}{2}\right)

, we get

y = \frac{1}{2\sin^2\left(\frac{x}{2}\right)} = \frac12 \csc^2\left(\frac{x}{2}\right)

Now integrate:

\int y \, dx = \frac12 \int \csc^2\left(\frac{x}{2}\right) dx

Let

u = \frac{x}{2}

, so

du = \frac{dx}{2}

i.e.

dx = 2\,du

Then integral becomes

\frac12 \int \csc^2 u \cdot 2\,du = \int \csc^2 u \, du = -\cot u + C

Substitute back

u = \frac{x}{2}

to obtain

-\cot\left(\frac{x}{2}\right) + C

Thus,

\int y \, dx = -\cot\left(\frac{x}{2}\right) + C

Answer:

Option B:

-\cot(x/2)+c

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