<s>[INST] Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is
$$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$
Is this true? If so, how to prove it?
 [/INST] Here I present a completely elementary proof using only high school level techniques.
With the obvious substitution $\displaystyle \large t = \tan{x}$
We proceed as follows: 
Using the Euler Substitution of the third kind, the following substitution is obtained:
$\displaystyle \large \sqrt{t-t^2} = tz \Rightarrow t = \frac{1}{1+z^2}$
$\displaystyle \large \text{d}t = \frac{-2z \, \text{d}z}{(1+z^2)^2}$ 
The borders change from $\displaystyle \large (0 \to 1)_t$ to $\displaystyle \large (\infty \to 0)_z$, and note the negative from the differential element flips this back into more "sensible" borders
The computation of the transform is tedious algebra and I will omit said details.
$\displaystyle \large \int_0^{\frac{\pi}{4}} \sqrt{\tan{x}}\sqrt{1-\tan{x}} \text{d}x = 2 \int_0^\infty \frac{z^2 \, \text{d}z}{(z^4+2z^2+2)(z^2+1)}$
It is easy to show that:
$\displaystyle \large \frac{z^2}{(z^4+2z^2+2)(1+z^2)} \equiv \frac{z^2+2}{z^4+2z^2+2} - \frac{1}{1+z^2}$
Simplifying:
$\displaystyle \large \int_0^{\frac{\pi}{4}} \sqrt{\tan{x}}\sqrt{1-\tan{x}} \text{d}x = 2 \int_0^\infty \frac{(z^2 + 2) \, \text{d}z}{z^4+2z^2+2} - \pi$
With the simultaneous substitutions $\displaystyle \large z = \frac{\sqrt[4]{2}}{u}$ and $\displaystyle \large z = \sqrt[4]{2}u$, and averaging, the integral is reduced to:
$\displaystyle \large \frac{1+\sqrt{2}}{\sqrt[4]{2}} \int_0^\infty \frac{(u^2+1) \, \text{d}u}{u^4+\sqrt{2}u^2+1}$
The substitution $\displaystyle \large v = u - \frac{1}{u}$ bijects the interval $\displaystyle \large (0,\infty)_u \mapsto (-\infty, \infty)_v$ and the integral is reduced to:
$\displaystyle \large \frac{1+\sqrt{2}}{\sqrt[4]{2}} \int_{-\infty}^\infty \frac{\text{d}v}{v^2+2+\sqrt{2}} - \pi = \pi \left(\frac{1+\sqrt{2}}{\sqrt{2(1+\sqrt{2})}} - 1 \right) = \pi \left(\sqrt{\frac{1+\sqrt{2}}{2}} - 1\right)$ 
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