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UPSC NDA I 2017 Mathematics Solved Paper

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Question : 58 of 120

Marks: +1, -0

What is

∫_{0}^{\frac{\pi }{2}}\frac{d\theta }{1+cos\theta }

equal to?

$\frac{1}{2}$
1
√3
None of the above

Solution:

Concept:

Use the identity

1+\cos\theta = 2\cos^2(\theta/2)

to simplify and integrate

\sec^2(\theta/2)

.

Explanation:

Let

I = \int_{0}^{\pi/2} \frac{d\theta}{1+\cos\theta}

.

Apply

1+\cos\theta = 2\cos^2(\theta/2)

.

Then

I = \int_{0}^{\pi/2} \frac{d\theta}{2\cos^2(\theta/2)} = \frac{1}{2} \int_{0}^{\pi/2} \sec^2(\theta/2) \, d\theta

.

Substitute

t = \theta/2

, so

d\theta = 2\,dt

.

Update limits: when

\theta=0

,

t=0

; when

\theta=\pi/2

,

t=\pi/4

.

Thus

I = \frac{1}{2} \int_{0}^{\pi/4} \sec^2 t \cdot 2\,dt = \int_{0}^{\pi/4} \sec^2 t \, dt

.

Integrate:

\int \sec^2 t \, dt = \tan t

.

Evaluate:

I = [\tan t]_{0}^{\pi/4} = \tan(\pi/4) - \tan(0) = 1 - 0 = 1

.

Answer:

1

(Option B)

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