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UPSC NDA I 2015 Mathematics Paper

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Question Numbers: 87-90

Direction: Consider the integral:

I_{m}=∫_{0}^{\pi }\frac{sin2mx}{sinx}dx,

where m is a positive integer.

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Question : 89 of 120

Marks: +1, -0

What is

I_{m}

is equal to?

0
1
m
2 m

Solution:

Concept:

Use the property of definite integrals:

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

.

Explanation:

Let

I_m = \int\limits_{0}^{\pi} \frac{\sin(2mx)}{\sin x} \, dx

.

Apply the property with

a=0

,

b=\pi

, so

f(x) = \frac{\sin(2mx)}{\sin x}

.

Then

f(\pi-x) = \frac{\sin(2m(\pi-x))}{\sin(\pi-x)}

.

Since

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

, and

\sin(2m(\pi-x)) = \sin(2m\pi - 2mx) = -\sin(2mx)

(because

\sin(2m\pi)=0

,

\cos(2m\pi)=1

), we get

f(\pi-x) = -\frac{\sin(2mx)}{\sin x} = -f(x)

.

Thus

I_m = \int\limits_{0}^{\pi} f(\pi-x) \, dx = \int\limits_{0}^{\pi} -f(x) \, dx = -I_m

.

Adding

I_m

to both sides gives

2I_m = 0

, so

I_m = 0

.

Answer:

0

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