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प्रश्न
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उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{\tan x}} d x ................(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{\tan\left( \frac{\pi}{2} - x \right)}} dx.....................\left[\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx\right]\]
\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{cotx}} d x .................(2)\]
\[\text{Adding (1) and (2) we get}\]
\[2I = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{cotx}} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{1 + \sqrt{cotx} + 1 + \sqrt{\tan x}}{\left( 1 + \sqrt{\tan x} \right) \left( 1 + \sqrt{cotx} \right)} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{1 + \sqrt{cotx} + 1 + \sqrt{\tan x}}{1 + \sqrt{cotx} + \sqrt{\tan x} + \sqrt{\tan x \ cotx}} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} dx\]
\[ = \int_0^\frac{\pi}{2} dx \]
\[ = \left[ x \right]_0^\frac{\pi}{2} \]
\[ = \frac{\pi}{2}\]
\[Hence\ I = \frac{\pi}{4}\]
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