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UPSC NDA 2 2022 Mathematics Solved Paper

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Question : 95 of 120

Marks: +1, -0

If

\displaystyle\int\limits_{0}^{\frac{\pi}{2}} (\sin^{4} x + \cos^{4} x) \, dx = k

, then what is the value of

\displaystyle\int\limits_{0}^{20\pi} (\sin^{4} x + \cos^{4} x) \, dx

?

k
10 k
20 k
40 k

Solution:

Concept:

If

f(x)

is periodic with period

T

, then

\int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx

.

Explanation:

Simplify

f(x) = \sin^4 x + \cos^4 x

.

Use the identity:

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

.

Rewrite

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

.

Thus

f(x) = 1 - \frac{1}{2} \cdot \frac{1 - \cos 4x}{2} = \frac{3}{4} + \frac{\cos 4x}{4}

.

Since

\cos 4x

has period

\frac{2\pi}{4} = \frac{\pi}{2}

,

f(x)

also has period

\frac{\pi}{2}

.

Given

\int_{0}^{\pi/2} f(x) \, dx = k

.

Now

\int_{0}^{20\pi} f(x) \, dx = \int_{0}^{40 \cdot \pi/2} f(x) \, dx

.

Using the periodic property, this equals

40 \int_{0}^{\pi/2} f(x) \, dx = 40k

.

Answer:

40k

(Option D)

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