So Small!

Write $\sqrt{\tan{\theta} }$ as

$\dfrac{ ( \sqrt{\tan{\theta}} +\sqrt{\cot{\theta}} ) + (\sqrt{\tan{\theta}} - \sqrt{\cot{\theta}} ) }{2}$

Consider $I_1 = \displaystyle \int_0^{\large\frac{\pi}{4}}\mathrm{ \frac{\sqrt{\cot{\theta}} + \sqrt{\tan{\theta}}~d\theta}{2}} = \displaystyle \int_0^{\large\frac{\pi}{2} }\mathrm{\frac{\sin{\theta} +\cos{\theta}~d\theta}{2\sqrt{\sin{\theta}\cos{\theta}}}}$

Substitute $\sin{\theta} - \cos{\theta} = t$ The numerator is $dt$ .The denominator can be easily expressed in terms of $t$

$( \displaystyle \text{Hint } : \sin{\theta}\cos{\theta} = \frac{ 1- t^2}{2})$

Similarly for $I_2$ = $\displaystyle \int_0^{\frac{\pi}{4}}\mathrm{\frac{\sqrt{\tan{\theta}} - \sqrt{\cot{\theta}} ~d\theta}{2}} = \int_0^{\frac{\pi}{4}}\mathrm{\frac{\cos{\theta} - \sin{\theta}~d\theta}{2\sqrt{\sin{\theta}\cos{\theta}}}}$ , substitute $\sin{\theta} + \cos{\theta} = m$

Required integral is $\color{#D61F06}{I_1 + I_2}$

This comes out to be

$\dfrac{ \ln{(\sqrt{2}-1)}}{\sqrt{2}} + \dfrac{\pi}{2\sqrt{2}} = \boxed{0.487495}$

With the substitution $v^2 = \tan x$ we obtain $\begin{aligned} \int_0^{\frac14\pi} \sqrt{\tan x}\,dx & = 2\int_0^1 \frac{v^2\,dv}{1+v^4} \; = \; \frac{1}{\sqrt{2}}\int_0^1 \left(\frac{v}{v^2 - \sqrt{2}v + 1} - \frac{v}{v^2 + \sqrt{2}v +1}\right)\,dv \\ & = \frac{1}{2\sqrt{2}}\Big[\ln\big(v^2 - \sqrt{2}v + 1\big) + 2\tan^{-1}\big(\sqrt{2}v - 1\big) - \ln\big(v^2 + \sqrt{2}v + 1\big) - 2\tan^{-1}\big(\sqrt{2}v + 1\big) \Big] _0^1 \\ & = \frac{1}{2\sqrt{2}}\Big\{ \ln\left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right) + 2\big[\tan^{-1}(\sqrt{2}-1) + \tan^{-1}(\sqrt{2}+1)\big]\Big\} \\ & = \frac{1}{2\sqrt{2}} \Big[2\ln(\sqrt{2}-1) + 2 \times \tfrac12\pi\Big] \\ & = \tfrac{1}{\sqrt{2}}\ln(\sqrt{2}-1) + \tfrac{1}{2\sqrt{2}}\pi \; = \; \boxed{0.487495} \end{aligned}$