UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 61-62

Consider the following for the next items that follow:

Let

I=∫_{-2\pi }^{2\pi }\frac{sin^{4}x+cos^{4}x}{1+3^{x}}dx

Question : 62 of 120

Marks: +1, -0

What is I equal to?

Solution:

Concept:

The definite integral property

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

is used. Trigonometric identities simplify

sin^4 x + cos^4 x

to a form easy to integrate.

Explanation:

Let

I = ∫_{-2π}^{2π} \frac{\sin^4 x + \cos^4 x}{1+3^x} \, dx

Using the property, replace

x

( -2π + 2π - x ) = -x

, giving

I = ∫_{-2π}^{2π} \frac{\sin^4 (-x) + \cos^4 (-x)}{1+3^{-x}} \, dx = ∫_{-2π}^{2π} \frac{\sin^4 x + \cos^4 x}{1+3^{-x}} \, dx

Adding the two expressions for

I

2I = ∫_{-2π}^{2π} (\sin^4 x + \cos^4 x) \left( \frac{1}{1+3^x} + \frac{1}{1+3^{-x}} \right) dx

Note

\frac{1}{1+3^x} + \frac{1}{1+3^{-x}} = \frac{1}{1+3^x} + \frac{3^x}{3^x+1} = 1

. So

2I = ∫_{-2π}^{2π} (\sin^4 x + \cos^4 x) dx

Rewrite

\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2 2x

Using

\sin^2 2x = \frac{1 - \cos 4x}{2}

, we get

1 - \frac{1}{4}(1 - \cos 4x) = \frac{3}{4} + \frac{\cos 4x}{4}

Thus

2I = ∫_{-2π}^{2π} \left( \frac{3}{4} + \frac{\cos 4x}{4} \right) dx = \left[ \frac{3}{4}x + \frac{\sin 4x}{16} \right]_{-2π}^{2π}

Evaluate:

\frac{3}{4}(2π - (-2π)) + \frac{1}{16}(\sin 8π - \sin (-8π)) = 3π + 0 = 3π

. Hence

I = \frac{3π}{2}

Answer:

I = \frac{3π}{2}

, option C.

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