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Π / 3 ∫ π / 6 1 1 + √ Tan X D X - Mathematics

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प्रश्न

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]
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उत्तर

\[Let\ I = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan x}} d x .................(1)\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{\tan\left( \frac{\pi}{3} + \frac{\pi}{6} - x \right)}} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{1 + \sqrt{cotx}}dx ....................(2)\]
\[\text{Adding (1) and (2)}\]
\[2I = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{cotx}} \right) d x \]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \frac{1 + \sqrt{cotx} + 1 + \sqrt{\tan x}}{1 + \sqrt{cotx} + \sqrt{\tan x} + \sqrt{\tan x cotx}} \right) dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} dx = \left[ x \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{\pi}{3} - \frac{\pi}{6}\]
\[ \therefore 2I = \frac{\pi}{6}\]
\[Hence\ I = \frac{\pi}{12}\]

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Definite Integrals
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अध्याय 20: Definite Integrals - Exercise 20.4 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.4 | Q 15 | पृष्ठ ६१

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