Question

\( \int_0^{\pi/2} \sqrt{\tan x} \, dx = \)

• A. \( \frac{\pi}{\sqrt{2}} \)
• B. \( \frac{\pi}{2} \)
• C. \( \sqrt{2}\pi \)
• D. \( 2\pi \)

Hint:

The integral \( \int_0^{\pi/2} \sqrt{\tan x} \, dx \) and \( \int_0^{\pi/2} \sqrt{\cot x} \, dx \) are classic problems. A key step is to add them together. The substitution \(t = \sin x - \cos x\) is crucial for solving the resulting integral. Both integrals evaluate to \( \frac{\pi}{\sqrt{2}} \).

Answer 1

The Correct Option is A

This is a standard special integral related to the Beta and Gamma functions, but it can be solved with substitution.
Let \( I = \int_0^{\pi/2} \sqrt{\tan x} \, dx \).
Using the property \( \int_0^a f(x)dx = \int_0^a f(a-x)dx \):
\( I = \int_0^{\pi/2} \sqrt{\tan(\pi/2 - x)} \, dx = \int_0^{\pi/2} \sqrt{\cot x} \, dx \).
Adding the two expressions for I:
\( 2I = \int_0^{\pi/2} (\sqrt{\tan x} + \sqrt{\cot x}) \, dx = \int_0^{\pi/2} \left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right) dx \).
\( 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi/2} \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} \, dx = \sqrt{2} \int_0^{\pi/2} \frac{\sin x + \cos x}{\sqrt{\sin(2x)}} \, dx \).
Now, we use the identity \( \sin(2x) = 1 - (\sin x - \cos x)^2 \).
Let \( t = \sin x - \cos x \). Then \( dt = (\cos x + \sin x) dx \).
Let's check the limits of integration for t.
When \(x=0\), \(t = \sin 0 - \cos 0 = -1\).
When \(x=\pi/2\), \(t = \sin(\pi/2) - \cos(\pi/2) = 1\).
The integral becomes \( 2I = \sqrt{2} \int_{-1}^1 \frac{dt}{\sqrt{1-t^2}} \).
This is a standard integral: \( \sqrt{2} [\sin^{-1} t]_{-1}^1 \).
\( 2I = \sqrt{2} (\sin^{-1}(1) - \sin^{-1}(-1)) = \sqrt{2}(\frac{\pi}{2} - (-\frac{\pi}{2})) = \sqrt{2}(\pi) \).
So, \( 2I = \pi\sqrt{2} \).
Therefore, \( I = \frac{\pi\sqrt{2}}{2} = \frac{\pi}{\sqrt{2}} \).