Question

$\int_0^{\pi/4} (\sqrt{\tan x} + \sqrt{\cot x}) dx =$

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

Hint:

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

Answer 1

The Correct Option is D

Step 1: Simplify Integrand
$\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} = \sqrt{2} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} = \sqrt{2} \frac{\sin x + \cos x}{\sqrt{1 - (\sin x - \cos x)^2}}$.
Step 2: Substitution
Let $\sin x - \cos x = t \implies (\cos x + \sin x) dx = dt$.
Limits: $x=0 \to t=-1$; $x=\pi/4 \to t=0$.
Step 3: Integration
$\sqrt{2} \int_{-1}^{0} \frac{dt}{\sqrt{1-t^2}} = \sqrt{2} [\sin^{-1} t]_{-1}^{0} = \sqrt{2} (0 - (-\pi/2)) = \frac{\pi}{\sqrt{2}}$.
Final Answer: (D)