0-pi-4-tan-x-1-tan-x-dx-

Question Number 129105 by I want to learn more last updated on 12/Jan/21

\int_{0}^{\frac{π}{4}} \sqrt{\tan x (1 - \tan x)} dx

Answered by Lordose last updated on 13/Jan/21

Ω = \int_{0}^{\frac{π}{4}} \sqrt{\tan (x)} \sqrt{1 - \tan (x)} dx \overset{u = \tan (x)}{=} \int_{0}^{1} \frac{u^{\frac{1}{2}} \sqrt{1 - u}}{1 + u^{2}} du

Ω = \int_{0}^{1} u^{\frac{1}{2}} {(1 - u)}^{\frac{1}{2}} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} u^{2 n} du = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} u^{2 n + \frac{1}{2}} {(1 - u)}^{\frac{1}{2}} du

β (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} dt

Ω = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (β (\frac{4 n + 3}{2}, \frac{3}{2})) = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\frac{Γ (\frac{4 n + 3}{2}) Γ (\frac{3}{2})}{Γ (2 n + 3)})

Ω = \frac{\sqrt{π}}{2} \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} (\frac{Γ (\frac{4 n + 3}{2})}{Γ (2 n + 3)}) = \frac{\sqrt{π}}{2} (\sqrt{π} (\sqrt{2 (1 + \sqrt{2})} - 2) = \frac{π}{2} (\sqrt{2 (1 + \sqrt{2})} - 2)

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Commented by MJS_new last updated on 13/Jan/21

great!

Commented by I want to learn more last updated on 13/Jan/21

Thanks sir .

Answered by MJS_new last updated on 13/Jan/21

\int \sqrt{(1 - \tan x) \tan x} d x =

[t = \sqrt{\frac{1}{\tan x} - 1} \to d x = - \frac{2 \sqrt{(1 - \tan x) \tan^{3} x}}{1 + \tan^{2} x} d t]

= - 2 \int \frac{t^{2}}{(t^{2} + 1) (t^{4} + 2 t^{2} + 2)} d t =

= - 2 \int \frac{t^{2}}{(t^{2} + 1) (t^{2} - \sqrt{- 2 + 2 \sqrt{2}} t + \sqrt{2}) (t^{2} + \sqrt{- 2 + 2 \sqrt{2}} t + \sqrt{2})} d t =

= - 2 \int \frac{t^{2}}{(t^{2} + 1) (t^{2} - a t + b) (t^{2} + a t + b)} d t =

= 2 \int \frac{d t}{t^{2} + 1} + \frac{\sqrt{- 2 + 2 \sqrt{2}}}{2} (\int \frac{t - 2 \sqrt{1 + \sqrt{2}}}{t^{2} - \sqrt{- 2 + 2 \sqrt{2}} t + \sqrt{2}} d t - \int \frac{t + 2 \sqrt{1 + \sqrt{2}}}{t^{2} + \sqrt{- 2 + 2 \sqrt{2}} t + \sqrt{2}} d t)

and these can be solved using the usual formulas

in the end I get

(\frac{\sqrt{2 + 2 \sqrt{2}}}{2} - 1) π

Commented by I want to learn more last updated on 13/Jan/21

Thanks sir .