Show that ∫_0 ^(π/4) (√(tan x)) (√(1−tan x)) dx=(((√((√2)−1))/( (√2)))−1)π


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Question Number 204275 by Frix last updated on 11/Feb/24

$Show that$ $\underset{0}{\int^{\frac{π}{4}}} \sqrt{\tan x} \sqrt{1 - \tan x} d x = (\frac{\sqrt{\sqrt{2} - 1}}{\sqrt{2}} - 1) π$
	Answered by witcher3 last updated on 11/Feb/24
	$nice problem tan(x)=y ∫_0 ^1 ((√(y(1−y)))/(1+y^2 ))dy=I;(1/y)=z⇒I=∫_1 ^∞ ((√(z−1))/(z(z^2 +1)))dz (√(z−1))=x⇒I=∫_0 ^∞ ((2x^2 )/(((x^2 +1)^2 +1)(x^2 +1)))dx =∫_0 ^∞ ((2x^2 )/((x^2 +1+i)(x^2 +1−i)))=∫_(−∞) ^∞ ((x^2 dx)/((x^2 +1−i)(x^2 +1+i)(x^2 +1)))=I Residue Theorem overH {z∈C∣im(z)≥0} x^2 +1−i=0⇒x^2 =(√2)e^(3((iπ)/4)) x1=(√(√2))e^((i3π)/8) ∈H x^2 =−1−i⇒x^2 =(√2)e^((5iπ)/4) ⇒x_2 =(√(√2))e^((5iπ)/8) x^2 +1=0⇒x_3 =i I=2iπ {Res(f,z_1 )+Res(f,z_2 ))Resf(x_3 )} =2iπ{(z_1 ^2 /((2z_(1 ) )(z_1 ^2 +1+i)(z_1 ^2 +1)))+(z_2 ^2 /((2z_2 )(z_2 ^2 +1−i)(z_2 ^2 +1)))+(z_3 ^2 /(2z_3 (z_3 ^2 +1+i)(z_3 ^2 +1−i)))) =2iπ{(((√(√2))e^((3iπ)/8) )/(4i^2 ))+(((√(√2))e^((5iπ)/8) )/(−4))+(i/(2(i)(−i)))} =(i/2)π(√(√2))(−e^((3iπ)/8) −e^((5iπ)/8) )−π =(π/2)(√(√2))(2sin(3(π/8)))−π =π(√((√2).sin^2 (((3π)/8))))−π =π(√((1/2)(1−cos(3(π/4)))(√2)))−π =π(√((1/( (√2)))+(1/2)))−π =π(((√((√2)+1))/( (√2)))−1)$
	$nice problem$ $\tan (x) = y$ $\int_{0}^{1} \frac{\sqrt{y (1 - y)}}{1 + y^{2}} dy = I; \frac{1}{y} = z \Rightarrow I = \int_{1}^{\infty} \frac{\sqrt{z - 1}}{z (z^{2} + 1)} dz$ $\sqrt{z - 1} = x \Rightarrow I = \int_{0}^{\infty} \frac{{2 x}^{2}}{({(x^{2} + 1)}^{2} + 1) (x^{2} + 1)} dx$ $= \int_{0}^{\infty} \frac{{2 x}^{2}}{(x^{2} + 1 + i) (x^{2} + 1 - i)} = \int_{- \infty}^{\infty} \frac{x^{2} dx}{(x^{2} + 1 - i) (x^{2} + 1 + i) (x^{2} + 1)} = I$ $Residue Theorem overH {z \in C ∣ im (z) ⩾ 0}$ $x^{2} + 1 - i = 0 \Rightarrow x^{2} = \sqrt{2} e^{3 \frac{i π}{4}}$ $x 1 = \sqrt{\sqrt{2}} e^{\frac{i 3 π}{8}} \in H$ $x^{2} = - 1 - i \Rightarrow x^{2} = \sqrt{2} e^{\frac{5 i π}{4}} \Rightarrow x_{2} = \sqrt{\sqrt{2}} e^{\frac{5 i π}{8}}$ $x^{2} + 1 = 0 \Rightarrow x_{3} = i$ $I = 2 i π {Res (f, z_{1}) + Res (f, z_{2})) Resf (x_{3})}$ $= 2 i π {\frac{z_{1}^{2}}{({2 z}_{1}) (z_{1}^{2} + 1 + i) (z_{1}^{2} + 1)} + \frac{z_{2}^{2}}{({2 z}_{2}) (z_{2}^{2} + 1 - i) (z_{2}^{2} + 1)} + \frac{z_{3}^{2}}{{2 z}_{3} (z_{3}^{2} + 1 + i) (z_{3}^{2} + 1 - i)})$ $= 2 i π {\frac{\sqrt{\sqrt{2}} e^{\frac{3 i π}{8}}}{{4 i}^{2}} + \frac{\sqrt{\sqrt{2}} e^{\frac{5 i π}{8}}}{- 4} + \frac{i}{2 (i) (- i)}}$ $= \frac{i}{2} π \sqrt{\sqrt{2}} (- e^{\frac{3 i π}{8}} - e^{\frac{5 i π}{8}}) - π$ $= \frac{π}{2} \sqrt{\sqrt{2}} (2 \sin (3 \frac{π}{8})) - π$ $= π \sqrt{\sqrt{2} . \sin^{2} (\frac{3 π}{8})} - π$ $= π \sqrt{\frac{1}{2} (1 - \cos (3 \frac{π}{4})) \sqrt{2}} - π$ $= π \sqrt{\frac{1}{\sqrt{2}} + \frac{1}{2}} - π$ $= π (\frac{\sqrt{\sqrt{2} + 1}}{\sqrt{2}} - 1)$
	Commented by Frix last updated on 11/Feb/24

	$Thank you!$
	Commented by witcher3 last updated on 11/Feb/24

	$withe pleasur$
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