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UPSC NDA 2 2019 Mathematics Solved Paper

Question Numbers: 94-95

Directions: Read the following information and answer the two items that follow:

Consider the integrals

I_{1}=\overset{\pi }{\underset{0}{∫}}\frac{xdx}{1+sinx} and I_{2}=\overset{\pi }{\underset{0}{∫}}\frac{(\pi -x)dx}{1-sin(\pi +x)}

Question : 95 of 120

Marks: +1, -0

What is the value of

I_{1} + I_{2}

Solution:

Concept:

Use the property of definite integrals:

\int\limits_{a}^{b} f(x) \, dx = \int\limits_{a}^{b} f(a+b-x) \, dx

Explanation:

Given:

I_{1} = \int\limits_{0}^{\pi} \frac{x \, dx}{1 + \sin x}

and

I_{2} = \int\limits_{0}^{\pi} \frac{(\pi - x) \, dx}{1 - \sin(\pi + x)}

Since

\sin(\pi + x) = -\sin x

, we have

1 - \sin(\pi + x) = 1 + \sin x

Thus

I_{2} = \int\limits_{0}^{\pi} \frac{(\pi - x) \, dx}{1 + \sin x}

Applying the property to

I_{1}

I_{1} = \int\limits_{0}^{\pi} \frac{(\pi - x) \, dx}{1 + \sin x} = I_{2}

I_{1} = I_{2}

Now write

I_{1}

in two forms:

(1)

I_{1} = \int\limits_{0}^{\pi} \frac{x \, dx}{1 + \sin x}

(2)

I_{1} = \int\limits_{0}^{\pi} \frac{(\pi - x) \, dx}{1 + \sin x}

Add these:

2I_{1} = \int\limits_{0}^{\pi} \frac{\pi \, dx}{1 + \sin x} = \pi \int\limits_{0}^{\pi} \frac{dx}{1 + \sin x}

Simplify

\frac{1}{1 + \sin x}

using the identity

\frac{1}{1 + \sin x} = \sec^2 x - \sec x \tan x

Thus

2I_{1} = \pi \int\limits_{0}^{\pi} (\sec^2 x - \sec x \tan x) \, dx = \pi \left[ \tan x - \sec x \right]_{0}^{\pi}

Evaluate:

\tan \pi = 0

\tan 0 = 0

;

\sec \pi = -1

\sec 0 = 1

2I_{1} = \pi \left[ (0 - (-1)) - (0 - 1) \right] = \pi (1 + 1) = 2\pi

Hence

I_{1} = \pi

Since

I_{2} = I_{1} = \pi

, we get

I_{1} + I_{2} = 2\pi

Answer:

I_{1} + I_{2} = 2\pi

(Option A).

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