$\int_0^{\frac{\pi}{4}}(\sqrt{\tan x}+\sqrt{\cot x}) \mathrm{d} x=$

Evaluate the integral: $I = \int_0^{\frac{\pi}{4}} (\sqrt{\tan x} + \sqrt{\cot x}) \, \mathrm{d}x$

Begin by simplifying the integrand:

$\sqrt{\tan x} + \sqrt{\cot x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} = \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}$

The integral becomes:

$I = \int_0^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, \mathrm{d}x$

Multiply numerator and denominator by $\sqrt{2}$ and recognize that $2\sin x \cos x = \sin 2x$:

$I = \int_0^{\frac{\pi}{4}} \frac{\sqrt{2}(\sin x + \cos x)}{\sqrt{\sin 2x}} \, \mathrm{d}x$

Use the substitution $t = \sin x - \cos x$, giving $\mathrm{d}t = (\cos x + \sin x) \, \mathrm{d}x$.
Then $t^2 = 1 - 2\sin x \cos x$, so $\sin x \cos x = \frac{1 - t^2}{2}$.

When $x = 0$, $t = -1$; when $x = \frac{\pi}{4}$, $t = 0$. Substituting:

$I = \int_{-1}^{0} \frac{\mathrm{d}t}{\sqrt{(1 - t^2)/2}} = \sqrt{2} \int_{-1}^{0} \frac{1}{\sqrt{1 - t^2}} \, \mathrm{d}t$

The antiderivative is $\arcsin t$, so:

$I = \sqrt{2} [\arcsin t]_{-1}^{0} = \sqrt{2} (\arcsin 0 - \arcsin(-1)) = \sqrt{2} \left(\frac{\pi}{2}\right)$

The final result is $I = \frac{\pi}{\sqrt{2}}$.