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UPSC NDA I 2014 Mathematics Paper

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For the next two (02) items that follow:

Consider the integrals

I_1=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}}

and

I_2=\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}

[NDA I 2014]

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Question : 87 of 120

Marks: +1, -0

What is

I_1 - I_2

0
$2I_1$
π
None of the above

Solution:

I_1 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}}

\Rightarrow I_1 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}} = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}}

\Rightarrow I_1+I_1 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\tan x}} + \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}} = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

\Rightarrow I_1 = \frac{\pi}{12}

I_2 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}\,dx}{\sqrt{\sin x}+\sqrt{\cos x}}

\Rightarrow I_2 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}\,dx}{\sqrt{\sin x}+\sqrt{\cos x}}

\Rightarrow I_2+I_2 = \int\limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

\Rightarrow I_1 = \frac{\pi}{12}

\Rightarrow I_1 - I_2 = 0

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