A Daunting Integral

Let $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} \,dx$

We make the substitution $u = \frac{\pi}{2} - x$ , which gives $I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cot x} \,dx$

Adding the two integrals together, we have

$2I = \int_{0}^{\frac{\pi}{2}} \sqrt{\tan x} + \sqrt{\cot x} \,dx = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \,dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{\cos x \sin x}} \,dx$

Now we make the substitution $u = \sin x - \cos x$ . We see that $du = ( \sin x + \cos x ) \, dx$ and $\cos x \sin x = \frac{1-u^{2}}{2}$ . Therefore,

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sqrt{\cos x \sin x}} \,dx = \int_{-1}^{1} \frac{\sqrt{2}}{\sqrt{1-u^{2}}} \,du = \sqrt{2} \arcsin(u) \big|_{-1}^{1} = \pi \sqrt{2}$

Then $2I = \pi \sqrt{2} \implies I = \boxed{\frac{\pi \sqrt{2}}{2}}$ .

$\int_0^{\frac12\pi} \sqrt{\tan x}\,dx \; = \; \int_0^{\frac12\pi} \sin^{\frac12}x \cos^{-\frac12}x\,dx \; = \; \tfrac12B(\tfrac34,\tfrac14) \; = \; \tfrac12\Gamma(\tfrac34)\Gamma(\tfrac14) \; = \; \tfrac12\pi \mathrm{cosec}\,\tfrac14\pi \; = \; \boxed{\tfrac{\pi}{\sqrt{2}}}$

Desmos exists.

After preforming the substitution $z=\sqrt{\tan{x}}$ the integral becomes $\int_{0}^{\infty} \frac{2z^2}{1+z^4} dz$ . Since $f(z)=\frac{z^2}{1+z^4}$ is an even function, $\int_{0}^{\infty} \frac{2z^2}{1+z^4} dz=\int_{-\infty}^{\infty}\frac{z^2}{1+z^4} dz$ . Now turning to complex analysis $\int_{-\infty}^{\infty} \frac{z^2}{1+z^4} dz \to \oint_\Gamma \frac{z^2}{1+z^4} dz$ as $R\to\infty$ where $\Gamma$ is taken to be the counterclockwise path along the interval $[-R,R]$ on the real axis and the semi-circle on the upper half plane centered at the origin with radius $R$ . This is because $f(z)$ is a meromorphic function on the complex plane and $\lim_{|z|\to\infty} \pi z f(z) = 0$ (Therefore the integral vanishes on the semi-circle part of the path due to the $ML$ Inequality). Now we can use the residue theorem and L'Hopital's rule to find that $\oint_\Gamma \frac{z^2}{1+z^4} dz = 2\pi i \sum_{i} res(f(z),z_i) = 2\pi i (\lim_{z\to e^{\frac{\pi i}{4}}} (z-e^{\frac{\pi i}{4}})f(z) + \lim_{z\to e^{\frac{3\pi i}{4}}} (z-e^{\frac{3\pi i}{4}})f(z))= \boxed{\frac{\pi}{\sqrt{2}}} \approx 2.221$ . (The points $z_i$ are the poles of $f(z)$ on the upper half of the complex plane).

Wolfram Alpha says

$\displaystyle \int_{0}^{\infty} \frac{2t^2}{1+t^4} dt = \int_{0}^{\infty} \frac{2}{1+t^4} dt$ (taking t=1/x)

$\displaystyle \int_{0}^{\infty} \frac{1+t^2}{1+t^4}dt$ (now dividing numerator and denominator by t² and taking x=t-1/t)

$\displaystyle \int_{-\infty}^{\infty} \frac{1}{x²+2}dx=\frac{\pi}{\sqrt2}$

Used wolfram alpha free widget for definite integral to get the answer.