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

Question Number 97569 by M±th+et+s last updated on 08/Jun/20

\int_{0}^{\frac{π}{4}} \sqrt{t a n (x)} \sqrt{1 - t a n (x)} d x

Answered by MJS last updated on 08/Jun/20

use thus :

t = \frac{\sqrt{1 - \tan x}}{\sqrt{\tan x}} \to d x = - {2 \cos}^{2} x \sqrt{\tan^{3} x} \sqrt{1 - \tan x} d t

\Rightarrow

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

now decompose \dots

Commented by MJS last updated on 08/Jun/20

t^{4} + 2 t^{2} + 2 = (t^{2} - \sqrt{2 \sqrt{2} - 2} t + \sqrt{2}) (t^{2} + \sqrt{2 \sqrt{2} - 2} t + \sqrt{2})

Commented by M±th+et+s last updated on 10/Jun/20

t h a n k y o u s i r m j s

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

s i m p l i f y i n g

\int_{0}^{\frac{π}{4}} \sqrt{t a n (x)} \sqrt{1 - t a n (x)} d x = 2 \int \frac{t^{2} + 2}{t^{4} + 2 t^{2} + 2} d t - π

w i t h t h e s i m u l t a n u s t = \frac{\sqrt[4]{2}}{u} a n d t = \sqrt[4]{2} u

a n d a v e r a g i n g, t h e i n t e g r a l r e d u c e d t o :

\frac{1 + \sqrt{2}}{\sqrt[4]{2}} \int_{0}^{\infty} \frac{(u^{2} + 1)}{u^{4} + \sqrt[4]{2} u^{2} + 1} d u

v = u - \frac{1}{u}

\int_{- \infty}^{\infty} \frac{d v}{v^{2} + 2 + \sqrt{2}} d v - π = π (\sqrt{\frac{1 + \sqrt{2}}{2}} - 1)