Question

Evaluate : \[ \int_{\frac{1}{12}}^{\frac{5}{12}} \frac{dx}{1+\sqrt{\cot x}} \]

Hint:

When an integral contains expressions like \(\sqrt{\tan x}\) or \(\sqrt{\cot x}\), try adding the integral with its complementary form using \(x \rightarrow \frac{\pi}{2}-x\). This often simplifies the expression.

Answer 1

Step 1: Use the complementary property of definite integrals.
Consider the function \[ I=\int_{a}^{b} \frac{dx}{1+\sqrt{\cot x}} \] Notice the identity \[ \cot\left(\frac{\pi}{2}-x\right)=\tan x \] Thus, \[ \sqrt{\cot\left(\frac{\pi}{2}-x\right)}=\sqrt{\tan x} \] Now define another integral \[ I=\int_{a}^{b}\frac{dx}{1+\sqrt{\tan x}} \] Step 2: Add the two forms.
Adding both expressions gives \[ 2I=\int_{a}^{b}\left(\frac{1}{1+\sqrt{\cot x}}+\frac{1}{1+\sqrt{\tan x}}\right)dx \] Now simplify the expression inside the bracket.
Let \[ t=\sqrt{\tan x} \] Then \[ \sqrt{\cot x}=\frac{1}{t} \] Thus \[ \frac{1}{1+1/t}+\frac{1}{1+t} \] Simplifying: \[ \frac{t}{t+1}+\frac{1}{t+1} \] \[ =\frac{t+1}{t+1} \] \[ =1 \] Hence \[ 2I=\int_{a}^{b}1\,dx \] Step 3: Evaluate the integral.
\[ 2I=b-a \] \[ 2I=\frac{5}{12}-\frac{1}{12} \] \[ 2I=\frac{4}{12} \] \[ 2I=\frac{1}{3} \] Thus \[ I=\frac{1}{6} \] Final Answer:
\[ \boxed{\frac{1}{6}} \]