`int_0^(pi//4) [sqrt((tan x)) + sqrt((cot x))] dx = `

Options

A. `pi/(sqrt2)`
B. `pi/(2sqrt(2))`
C. `(-(3 pi)/sqrt2)`
D. None

Answer

To solve the integral

\[ I = \int_0^{\frac{\pi}{4}} \left( \sqrt{\tan x} + \sqrt{\cot x} \right) dx, \]

we can start by rewriting \(\tan x\) and \(\cot x\) in terms of sine and cosine:

\[ I = \int_0^{\frac{\pi}{4}} \left( \sqrt{\frac{\sin x}{\cos x}} + \sqrt{\frac{\cos x}{\sin x}} \right) dx. \]

This can be simplified to:

\[ I = \int_0^{\frac{\pi}{4}} \left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right) dx. \]

Next, we can combine the two terms under a common denominator:

\[ I = \int_0^{\frac{\pi}{4}} \left( \frac{\sqrt{\sin^2 x} + \sqrt{\cos^2 x}}{\sqrt{\sin x \cos x}} \right) dx. \]

Now, we can express \(\sqrt{\sin^2 x} + \sqrt{\cos^2 x}\) as \(\sqrt{\sin^2 x + \cos^2 x}\) since \(\sin^2 x + \cos^2 x = 1\):

\[ I = \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\sin x \cos x}} dx. \]

Using the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\), we can rewrite the integral:

\[ I = \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\frac{1}{2} \sin(2x)}} dx = \sqrt{2} \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\sin(2x)}} dx. \]

Now, we can use the substitution \(u = 2x\), which gives \(du = 2dx\) or \(dx = \frac{du}{2}\). The limits change as follows: when \(x = 0\), \(u = 0\) and when \(x = \frac{\pi}{4}\), \(u = \frac{\pi}{2}\):

\[ I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}} \cdot \frac{du}{2} = \frac{\sqrt{2}}{2} \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}} du. \]

The integral \(\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}} du\) is a known integral and evaluates to \(\sqrt{\pi}\):

\[ I = \frac{\sqrt{2}}{2} \cdot \sqrt{\pi} = \frac{\sqrt{2\pi}}{2}. \]

Thus, the final result is:

\[ I = \frac{\sqrt{2\pi}}{2}. \]