Show-that-0-pi-4-tan-x-1-tan-x-dx-2-1-2-1-pi-

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

\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