The integral _ /6 ^ /4 dx ,2x , ( ^5 ,x + ^5 ,x ) equals - Vidyadip

MCQ

The integral $\int_{\pi /6}^{\pi /4} {\frac{{dx}}{{\sin \,2x\,\left( {{{\tan }^5}\,x + {{\cot }^5}\,x} \right)}}} $ equals

A
$\frac{1}{{20}}\,{\tan ^{ - 1}}\,\left( {\frac{1}{{9\sqrt 3 }}} \right)$
✓
$\frac{1}{{10}}\,\left( {\frac{\pi }{4} - {{\tan }^{ - 1}}\,\left( {\frac{1}{{9\sqrt {1\sqrt 3 } }}} \right)} \right)$
C
$\frac{\pi }{{40}}$
D
$\frac{1}{5}\,\left( {\frac{\pi }{4} - {{\tan }^{ - 1}}\,\left( {\frac{1}{{3\sqrt 3 }}} \right)} \right)$

✓

Answer

Correct option: B.

$\frac{1}{{10}}\,\left( {\frac{\pi }{4} - {{\tan }^{ - 1}}\,\left( {\frac{1}{{9\sqrt {1\sqrt 3 } }}} \right)} \right)$

b
$I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\frac{{{{\sec }^2}xdx}}{{2\tan x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $

$\text { Put } \tan x=t$

$ = \int\limits_{\frac{1}{{\sqrt 3 }}}^1 {\frac{{dt}}{{2t\left( {{t^5} + \frac{1}{{{t^5}}}} \right)}}} $

$ = \int\limits_{\frac{1}{{\sqrt 3 }}}^1 {\frac{{{t^4}dt}}{{2\left( {{t^{10}} + 1} \right)}}} $

$\text { Put } t^{5}=y$

$\left.\mathrm{I}=\frac{1}{10} \tan ^{-1}(\mathrm{y})\right]_{3}^{1} \frac{\mathrm{s}}{2}$

$=\frac{1}{10}\left(\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3^{5 / 2}}\right)\right)$

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