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UPSC NDA I 2024 Mathematics Paper

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Question : 78 of 120

Marks: +1, -0

What is

\rm \int_0^{\frac{\pi}{2}} \frac{a+\sin x}{2 a+\sin x+\cos x} d x

equal to ?

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

Solution:

Concept:

Use the property:

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

.

Explanation:

Let

I = \int_0^{\pi/2} \frac{a+\sin x}{2a+\sin x+\cos x}\,dx

.

Apply the property with

a=\pi/2

: replace

x

by

\pi/2 - x

.

Then

I = \int_0^{\pi/2} \frac{a+\sin(\pi/2 - x)}{2a+\sin(\pi/2 - x)+\cos(\pi/2 - x)}\,dx

.

Since

\sin(\pi/2 - x)=\cos x

and

\cos(\pi/2 - x)=\sin x

, we get:

I = \int_0^{\pi/2} \frac{a+\cos x}{2a+\cos x+\sin x}\,dx

.

Add this expression for

I

to the original

I

:

2I = \int_0^{\pi/2} \frac{(a+\sin x)+(a+\cos x)}{2a+\sin x+\cos x}\,dx = \int_0^{\pi/2} \frac{2a+\sin x+\cos x}{2a+\sin x+\cos x}\,dx

.

Simplify:

2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}

.

Thus

I = \frac{\pi}{4}

.

Answer:

\frac{\pi}{4}

(option A).

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