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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 76 of 120

Marks: +1, -0

What is

\int_0^{2\pi} \sqrt{1 + \sin \frac{x}{2}} \, \mathrm{d}x

equal to?

Solution:

Concept:

Use trigonometric identities to rewrite the integrand as a perfect square, then integrate with substitution and absolute value handling.

Explanation:

Set

1 = \sin^2(\frac{x}{4}) + \cos^2(\frac{x}{4})

and

\sin\frac{x}{2} = 2\sin(\frac{x}{4})\cos(\frac{x}{4})

The integrand becomes

\sqrt{\sin^2(\frac{x}{4}) + \cos^2(\frac{x}{4}) + 2\sin(\frac{x}{4})\cos(\frac{x}{4})} = \sqrt{(\sin(\frac{x}{4}) + \cos(\frac{x}{4}))^2}

This simplifies to

\left|\sin(\frac{x}{4}) + \cos(\frac{x}{4})\right|

Let

t = \frac{x}{4}

, so

dx = 4\,dt

. Limits:

x = 0 \to t = 0

;

x = 2\pi \to t = \frac{\pi}{2}

Then

I = 4\int_0^{\pi/2} \left|\sin t + \cos t\right|\,dt

[0, \frac{\pi}{2}]

\sin t + \cos t = \sqrt{2}\sin(t+\frac{\pi}{4}) \ge 0

, so we drop the absolute value.

Integrate:

\int (\sin t + \cos t)\,dt = -\cos t + \sin t

Evaluate:

4\left[(-\cos t + \sin t)\right]_0^{\pi/2} = 4\left[(-\cos\frac{\pi}{2}+\sin\frac{\pi}{2}) - (-\cos 0 + \sin 0)\right] = 4\left[(0+1) - (-1+0)\right] = 4(2) = 8

Answer:

8

(Option A).

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