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NDA Math Definite Integrals

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Consider the integrals

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

and

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

[NDA I 2014]

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Question : 61 of 62

Marks: +1, -0

What is

I_1 - I_2

0
$2I_1$
π
None of the above

Solution:

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

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

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

\Rightarrow I_1 = \pi/12

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

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

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

\Rightarrow I_1 = \pi/12

\Rightarrow I_1 - I_2 = 0

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