Question

Evaluate the integral: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot x}} \, dx = \]

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{\pi}{12} \)

Hint:

Use the property \( \int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx \) to simplify symmetric integrals, especially those involving complementary trigonometric identities.

Answer 1

The Correct Option is D

Let the given integral be: \[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot x}} \, dx \]
We use the property of definite integrals: \[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
Here, \( a = \frac{\pi}{6}, b = \frac{\pi}{3} \Rightarrow a + b = \frac{\pi}{2} \). So,
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\cot\left(\frac{\pi}{2} - x\right)}} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1 + \sqrt{\tan x}} \, dx \]
Now add the two forms: \[ 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{1}{1 + \sqrt{\cot x}} + \frac{1}{1 + \sqrt{\tan x}} \right) dx \]
Let \( A = \frac{1}{1 + \sqrt{\cot x}} + \frac{1}{1 + \sqrt{\tan x}} \).
Let \( \sqrt{\cot x} = a \Rightarrow \sqrt{\tan x} = \frac{1}{a} \), since \( \tan x = \frac{1}{\cot x} \)
Then: \[ A = \frac{1}{1 + a} + \frac{1}{1 + \frac{1}{a}} = \frac{1}{1 + a} + \frac{a}{a + 1} = \frac{1 + a}{1 + a} = 1 \]
So, \( 2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 \, dx = \left[ x \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \Rightarrow I = \frac{\pi}{12} \)