\(\int_{0}^{\pi / 2}(\sqrt{\tan x}+\sqrt{\cot x}) d x=\)

Calculation

I = \(\int_{0}^{\frac{\pi}{2}} (\sqrt{\tan x} + \sqrt{\cot x}) dx\)

⇒ I = \(\int_{0}^{\frac{\pi}{2}} (\sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}}) dx\)

⇒ I = \(\int_{0}^{\frac{\pi}{2}} (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}) dx\)

Let z = sin x - cos x, dz = (cos x + sin x) dx

z² = sin²x + cos²x - 2sin x cos x

⇒ z² = 1 - 2sin x cos x

⇒ 2sin x cos x = 1 - z²

⇒ sin x cos x = \(\frac{1-z^2}{2}\)

When x = 0, z = -1; when x = \(\frac{\pi}{2}\), z = 1

I = \(\int_{-1}^{1} \frac{\sqrt{2}}{\sqrt{1-z^2}} dz\)

⇒ I = \([\sqrt{2} \sin^{-1}z]_{-1}^{1}\)

⇒ I = \(\sqrt{2} (\frac{\pi}{2} - (-\frac{\pi}{2}))\)

⇒ I = \(\sqrt{2} \pi\)

Hence option 2 is correct