What is the value of I₂ ?

Correct Answer is : π2{π}{{2}}2π

Correct Answer is : π2{π}{{2}}2π

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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 81-83

Consider the following for the next three (03) items that follow :

Let

I_{1}=∫_{0}^{\pi }\frac{x}{1+cos^{2}x}dx

and

I_{2}=∫_{0}^{\pi }\frac{1}{1+sin^{2}x}dx

Question : 83 of 120

Marks: +1, -0

What is the value of

I_{2}

Solution:

Concept:

Use integration by substitution and the property

\int\limits_{0}^{\pi} f(x) \, dx = 2 \int\limits_{0}^{\pi/2} f(x) \, dx

when

f(x)=f(\pi-x)

Explanation:

We have

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

Since

\sin^{2}x = \sin^{2}(\pi-x)

, we can write

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

Divide numerator and denominator by

\cos^{2}x

to get

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

Let

t = \tan x

, then

dt = \sec^{2}x \, dx

When

x=0

t=0

; when

x=\pi/2

t \to \infty

Thus,

I_{2} = 2 \int\limits_{0}^{\infty} \frac{dt}{1+2t^{2}} = 2 \cdot \frac{1}{\sqrt{2}} \left[ \tan^{-1}(t\sqrt{2}) \right]_{0}^{\infty}

Using

\tan^{-1}(\infty) = \pi/2

, we get

I_{2} = \frac{2}{\sqrt{2}} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}

Answer:

\frac{\pi}{\sqrt{2}}

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