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UPSC NDA I 2024 Mathematics Paper

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Question : 74 of 120

Marks: +1, -0

What is

\rm \int_{-1}^1(3 \sin x-\sin 3 x) \cos ^2 x d x

equal to ?

$-\frac{1}{4}$
0
$\frac{1}{2}$
$\frac{1}{4}$

Solution:

Concept:

If

f(x)

is an odd function, i.e.,

f(-x) = -f(x)

, then the integral over symmetric limits is zero:

\int_{-a}^{a} f(x) \, dx = 0

.

Explanation:

Let

I = \int_{-1}^{1} (3 \sin x - \sin 3x) \cos^2 x \, dx

.

Define

f(x) = (3 \sin x - \sin 3x) \cos^2 x

.

Compute

f(-x)

:

f(-x) = (3 \sin(-x) - \sin(-3x)) \cos^2(-x)

.

Using

\sin(-x) = -\sin x

and

\cos(-x) = \cos x

, we get

f(-x) = (-3 \sin x + \sin 3x) \cos^2 x

.

Factor

-1

:

f(-x) = -(3 \sin x - \sin 3x) \cos^2 x = -f(x)

.

Thus

f(x)

is an odd function.

Hence

I = \int_{-1}^{1} f(x) \, dx = 0

.

Answer:

0, which corresponds to option B.

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