Question

plz solve this question..​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \int\limits_{0}^{1} \frac{1 - {x}^{2} }{1 + {x}^{2} } \, \frac{dx}{ \sqrt{1 + {x}^{4} } } \: \: = \frac{\pi}{k \sqrt{2} } \: }$

TO DETERMINE

The value of k

EVALUATION

Here

$\displaystyle \sf{ \int\limits_{0}^{1} \frac{1 - {x}^{2} }{1 + {x}^{2} } \, \frac{dx}{ \sqrt{1 + {x}^{4} } } \: \: } \:$

$= \displaystyle \sf{ \int\limits_{0}^{1} \frac{ \frac{1}{ {x}^{2} } - 1 }{x + \frac{1}{x} } \, \frac{dx}{ \sqrt{ {x}^{2} + \frac{1}{ {x}^{2} } } } \: \: } \:$

$= \displaystyle \sf{ \int\limits_{0}^{1} \frac{ \frac{1}{ {x}^{2} } - 1 }{x + \frac{1}{x} } \, \frac{dx}{ \sqrt{( {x + \frac{1}{x} )}^{2} - 2.x. \frac{1}{x} } } \: \: } \:$

$= \displaystyle \sf{ \int\limits_{0}^{1} \frac{ \frac{1}{ {x}^{2} } - 1 }{x + \frac{1}{x} } \, \frac{dx}{ \sqrt{( {x + \frac{1}{x} )}^{2} - 2 } } \: \: } \:$

$\displaystyle \sf{ x + \frac{1}{x} = \sqrt{2} \sec \theta}$

Then

$\displaystyle \sf{1 - \frac{1}{ {x}^{2} } = \sqrt{2} \sec \theta \tan \theta }$

$\displaystyle \sf{ Also \: when \: \: x = 0 \: , \theta = \frac{\pi}{2} }$

And

$\displaystyle \sf{when \: \: x = 1 \: , \theta = \frac{\pi}{4} }$

So the given Integral

$= \displaystyle \sf{ \int\limits_{0}^{1} \frac{ \frac{1}{ {x}^{2} } - 1 }{x + \frac{1}{x} } \, \frac{dx}{ \sqrt{( {x + \frac{1}{x} )}^{2} - 2 } } \: \: } \:$

$= \displaystyle \sf{ \int\limits_{ \frac{\pi}{2} }^{ \frac{\pi}{4} } \frac{ - \sqrt{2} \sec \theta \tan \theta }{\sqrt{2} \sec \theta } \, \frac{d \theta}{\sqrt{2} \tan \theta } \: \: } \:$

$= \displaystyle \sf{ - \frac{1}{ \sqrt{2} } \int\limits_{ \frac{\pi}{2} }^{ \frac{\pi}{4} } \, d\theta } \:$

$= \displaystyle \sf{ \frac{1}{ \sqrt{2} } \int\limits_{ \frac{\pi}{4} }^{ \frac{\pi}{2} } \, d\theta } \:$

$= \displaystyle \sf{ \frac{\pi}{ 4\sqrt{2} } } \:$

Now it is given that

$\displaystyle \sf{ \int\limits_{0}^{1} \frac{1 - {x}^{2} }{1 + {x}^{2} } \, \frac{dx}{ \sqrt{1 + {x}^{4} } } \: \: = \frac{\pi}{k \sqrt{2} } \: }$

$\implies \: \displaystyle \sf{ \frac{\pi}{k \sqrt{2} } = \frac{\pi}{4 \sqrt{2} } \: }$

Comparing we get

$\sf{ k = 4\: }$

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