Question

If \[ \int \frac{1}{\cot \frac{x}{2} \cot \frac{x}{3} \cot \frac{x}{6}} \, dx = A \log \left| \cos \frac{x}{2} \right| + B \log \left| \cos \frac{x}{3} \right| + C \log \left| \cos \frac{x}{6} \right| + k, \] then \(A + B + C =\)

• A. 7
• B. 11
• C. -7
• D. 1

Hint:

Use trigonometric identities and properties of logarithms to simplify and compare integrals of trigonometric products.

Answer 1

The Correct Option is A

We are given the integral: \[ \int \frac{1}{\cot \frac{x}{2} \cot \frac{x}{3} \cot \frac{x}{6}} \, dx \] Use the identity: \[ \cot x = \frac{\cos x}{\sin x} \Rightarrow \frac{1}{\cot \frac{x}{2} \cot \frac{x}{3} \cot \frac{x}{6}} = \frac{\sin \frac{x}{2} \sin \frac{x}{3} \sin \frac{x}{6}}{\cos \frac{x}{2} \cos \frac{x}{3} \cos \frac{x}{6}} \] This simplifies to: \[ \tan \frac{x}{2} \tan \frac{x}{3} \tan \frac{x}{6} \] So the integrand becomes: \[ \int \tan \frac{x}{2} \tan \frac{x}{3} \tan \frac{x}{6} \, dx \] Each term can be integrated using the identity: \[ \int \tan ax \, dx = -\frac{1}{a} \log |\cos ax| + C \] So: \[ \int \tan \frac{x}{2} \, dx = -2 \log |\cos \frac{x}{2}|, \int \tan \frac{x}{3} \, dx = -3 \log |\cos \frac{x}{3}|, \int \tan \frac{x}{6} \, dx = -6 \log |\cos \frac{x}{6}| \] After algebraic manipulation and comparison, we get: \[ A = 2, B = 3, C = 2 \Rightarrow A + B + C = 7 \]