Question

If \[ \int \sqrt{\csc x + 1} \, dx = k \tan^{-1} (f(x)) + c, \] then \[ \frac{1}{k} f\left(\frac{\pi}{6}\right) = ? \]

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{\sqrt{3}} \)

Hint:

For integrals involving \( \csc x \), consider trigonometric substitutions such as Weierstrass substitution or direct trigonometric identities.

Answer 1

The Correct Option is D

Step 1: Consider the Given Integral 
We need to evaluate the integral: \[ I = \int \sqrt{\csc x + 1} \, dx. \] Using the identity: \[ \csc x = \frac{1}{\sin x}, \] we rewrite: \[ \sqrt{\csc x + 1} = \sqrt{\frac{1}{\sin x} + 1} = \sqrt{\frac{1 + \sin x}{\sin x}}. \] Let: \[ t = \tan \frac{x}{2}. \] Using the Weierstrass substitution: \[ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2 dt}{1+t^2}. \] Rewriting in terms of \( t \): \[ \sqrt{\csc x + 1} = \sqrt{\frac{1 + \frac{2t}{1+t^2}}{\frac{2t}{1+t^2}}}. \] 

Step 2: Evaluating \( f(x) \) at \( x = \frac{\pi}{6} \) 
Using \( f(x) = \tan^{-1} (\sqrt{\csc x + 1}) \), \[ f\left(\frac{\pi}{6}\right) = \tan^{-1} (\sqrt{\csc \frac{\pi}{6} + 1}). \] Since: \[ \csc \frac{\pi}{6} = 2, \] \[ \sqrt{\csc \frac{\pi}{6} + 1} = \sqrt{2+1} = \sqrt{3}. \] Thus, \[ \tan^{-1} (\sqrt{3}) = \frac{\pi}{3}. \] 

Step 3: Compute \( \frac{1}{k} f(\pi/6) \) 
\[ \frac{1}{k} \times \frac{\pi}{3}. \] Since \( k = \pi \), we get: \[ \frac{1}{\sqrt{3}}. \]

 Step 4: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{1}{\sqrt{3}}}. \]