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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 : 94 of 120

Marks: +1, -0

What is the value of

I_{1}

Solution:

Concept:

Use the property of definite integrals:

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

, and standard integrals

\int\sec^2 x \, dx = \tan x + C

\int\sec x \tan x \, dx = \sec x + C

Explanation:

Given:

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

and

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

, but we only need

I_1

Apply the property to

I_1

: replace

x

\pi - x

Then

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

, since

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

Add the original

I_1

and this expression:

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

Simplify

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

Thus

2I_{1} = \pi \int_{0}^{\pi} \left( \sec^2 x - \sec x \tan x \right) dx

Evaluate:

\int\sec^2 x \, dx = \tan x

\int\sec x \tan x \, dx = \sec x

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

Values:

\tan \pi = 0

\sec \pi = -1

\tan 0 = 0

\sec 0 = 1

Hence

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

Therefore

I_{1} = \pi

Answer:

\pi

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