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Question Number 118757 by mohammad17 last updated on 19/Oct/20

Commented by mohammad17 last updated on 19/Oct/20

$w i t h o u t x = t a n θ$
	Answered by mnjuly1970 last updated on 19/Oct/20

	$\frac{π}{4} \overset{?}{=} a n s w e r$ $m e t h o d 1 :$ $e u l e r r e f l e c t i o n f o r m u l a .$ $x^{2} = t \Rightarrow Ω = \frac{1}{2} \int_{0}^{\infty} \frac{t^{(\frac{- 1}{2} + 1) - 1}}{{(t + 1)}^{2}} d t$ $\Rightarrow Ω = \frac{1}{2} β (\frac{1}{2}, \frac{3}{2}) = \frac{1}{2} Γ (\frac{1}{2}) Γ (\frac{3}{2})$ $= {(\frac{1}{2})}^{2} Γ^{2} (\frac{1}{2}) \overset{Γ (\frac{1}{2}) = \sqrt{π}}{=} \frac{π}{4} ✓ ✓$ $. . . m . n . j u l y . 1970 . . .$ $. . . . . . . . . .$ $m e t h o t 2 :$ $x = \frac{1}{t} \Rightarrow Ω = \int_{0}^{\infty} \frac{d t}{t^{2} {(1 + \frac{1}{t^{2}})}^{2}}$ $= \int_{0}^{\infty} \frac{(t^{2} + 1 - 1)}{{(1 + t^{2})}^{2}} d t = \frac{π}{2} - Ω$ $2 Ω = \frac{π}{2} \Rightarrow Ω = \frac{π}{4} ✓ ✓$ $. m . n . 1970 . .$
	Commented by mohammad17 last updated on 19/Oct/20

	$y e s s i r i t s r i g h t$
	Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

	$excellent$
	Commented by mnjuly1970 last updated on 19/Oct/20

	$t h a n k y o u$
	Commented by mnjuly1970 last updated on 19/Oct/20

	$s i n c e r e l y y o u r s$
	Answered by 1549442205PVT last updated on 19/Oct/20

	$Put x = \tan φ \Rightarrow dx = (1 + \tan^{2} φ) d φ$ $\int_{0}^{\infty} \frac{dx}{{(x^{2} + 1)}^{2}} = \int_{0}^{\frac{π}{2}} \frac{d φ}{(1 + \tan^{2} φ)} = \int_{0}^{\frac{π}{2}} \cos^{2} φ d φ$ $= \int_{0}^{\frac{π}{2}} \frac{1 + \cos 2 φ}{2} d φ = {(\frac{φ}{2} + \frac{1}{4} \sin 2 φ)}_{0}^{\frac{π}{2}}$ $= \frac{π}{4}$
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