Question

Evaluate \( \int \frac{dx}{1 + a \cos x} \), given \( a>|\sec \theta| \)

• A. \( \frac{1}{\sqrt{a^2 - 1}} \tan^{-1} \left( \frac{\sqrt{a - 1}}{\sqrt{a + 1}} \tan \frac{x}{2} \right) + C \)
• B. \( \frac{1}{\sqrt{a^2 - 1}} \tan^{-1} \left( \frac{\sqrt{1 - a}}{\sqrt{1 + a}} \tan \frac{x}{2} \right) + C \)
• C. Logarithmic form (incorrectly signed)
• D. \( \frac{1}{\sqrt{a^2 - 1}} \log \left( \frac{\sqrt{a + 1} \cos \frac{x}{2} + \sqrt{a - 1} \sin \frac{x}{2}}{\sqrt{a + 1} \cos \frac{x}{2} - \sqrt{a - 1} \sin \frac{x}{2}} \right) + C \)

Hint:

Use Weierstrass substitution or known forms for \( \int \frac{1}{1 + a \cos x} dx \) with \( a>1 \).

Answer 1

The Correct Option is D

We use standard integral: \[ \begin{align} \int \frac{dx}{1 + a \cos x} = \frac{1}{\sqrt{a^2 - 1}} \log \left( \frac{\sqrt{a + 1} \tan \left( \frac{x}{2} \right) + \sqrt{a - 1}}{\sqrt{a + 1} \tan \left( \frac{x}{2} \right) - \sqrt{a - 1}} \right) \] Using substitution \( \tan \frac{x}{2} = t \), and other trigonometric identities: Eventually, for \( a>1 \), it simplifies into: \[ \begin{align} \frac{1}{\sqrt{a^2 - 1}} \log \left( \frac{ \sqrt{a+1} \cos \frac{x}{2} + \sqrt{a - 1} \sin \frac{x}{2} }{ \sqrt{a+1} \cos \frac{x}{2} - \sqrt{a - 1} \sin \frac{x}{2} } \right) + C \]