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प्रश्न

\[\int\limits_0^{\pi/2} \frac{1}{1 + \sqrt{\tan x}} dx\]
योग

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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Definite Integrals
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अध्याय 20: Definite Integrals - Exercise 20.5 [पृष्ठ ९४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.5 | Q 6 | पृष्ठ ९४

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