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UPSC NDA I 2022 Mathematics Solved Paper

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Question : 61 of 120

Marks: +1, -0

What is

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

equal to?

$-\frac{1}{2}$
$\frac{1}{2}$
1
$\frac{3}{2}$

Solution:

Concept:

Use the substitution method and the identity

\sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)

or directly transform the integrand into

\sec^2 x / (1+\tan x)^2

to simplify the definite integral.

Explanation:

We start with

I = \int_{0}^{\pi/4} \frac{dx}{(\sin x + \cos x)^2}

.

Rewrite

\sin x = \frac{\tan x}{\sec x}

and

\cos x = \frac{1}{\sec x}

.

Then the denominator becomes

\left(\frac{\tan x}{\sec x} + \frac{1}{\sec x}\right)^2 = \frac{(\tan x + 1)^2}{\sec^2 x}

.

Thus

I = \int_{0}^{\pi/4} \frac{\sec^2 x \, dx}{(\tan x + 1)^2}

.

Now substitute

t = \tan x

, so

dt = \sec^2 x \, dx

.

When

x = 0

,

t = 0

; when

x = \pi/4

,

t = 1

.

The integral becomes

I = \int_{0}^{1} \frac{dt}{(t + 1)^2}

.

Let

v = t + 1

, then

dv = dt

.

New limits:

t=0 \Rightarrow v=1

;

t=1 \Rightarrow v=2

.

So

I = \int_{1}^{2} \frac{dv}{v^2}

.

Using

\int v^{-2} dv = -v^{-1}

, we get

I = \left[ -\frac{1}{v} \right]_{1}^{2} = -\frac{1}{2} - \left(-\frac{1}{1}\right) = \frac{1}{2}

.

Therefore

I = \frac{1}{2}

.

Answer:

\frac{1}{2}

(Option B).

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