Question

Evaluate the integral: \[ \int \frac{\sec x}{3(\sec x + \tan x) + 2} \,dx \]

• A. \( \frac{1}{2} \log \left| \frac{\tan \frac{x}{2} + 1}{\tan \frac{x}{2} + 5} \right| + C \)
• B. \( \frac{2}{\sqrt{11}} \tan^{-1} \left( \frac{3\tan \frac{x}{2} + 4}{\sqrt{11}} \right) + C \)
• C. \( \log |3 \sec x + 2 \tan x| + C \)
• D. \( \log |3 \tan x + 2 \sec x| + C \)

Hint:

The Weierstrass substitution \( t = \tan \frac{x}{2} \) helps simplify trigonometric integrals involving secant and tangent functions.

Answer 1

The Correct Option is A

Step 1: Substituting \( \tan \frac{x}{2} \) Using the Weierstrass substitution: \[ t = \tan \frac{x}{2} \] We apply the standard transformations: \[ \sec x = \frac{1 + t^2}{1 - t^2}, \quad \tan x = \frac{2t}{1 - t^2}, \quad dx = \frac{2 dt}{1 + t^2} \] 

Step 2: Expressing the Denominator \[ 3(\sec x + \tan x) + 2 \] \[ = 3 \left( \frac{1 + t^2}{1 - t^2} + \frac{2t}{1 - t^2} \right) + 2 \] \[ = 3 \cdot \frac{1 + t^2 + 2t}{1 - t^2} + 2 \] \[ = \frac{3 + 3t^2 + 6t}{1 - t^2} + 2 \] \[ = \frac{3 + 3t^2 + 6t + 2(1 - t^2)}{1 - t^2} \] \[ = \frac{3 + 3t^2 + 6t + 2 - 2t^2}{1 - t^2} \] \[ = \frac{1 + t^2 + 6t + 3}{1 - t^2} \] \[ = \frac{(\tan \frac{x}{2} + 5)(\tan \frac{x}{2} + 1)}{1 - t^2} \] 

Step 3: Integrating \[ \int \frac{\sec x}{(\tan \frac{x}{2} + 5)(\tan \frac{x}{2} + 1)} \,dx \] Applying logarithm integration properties: \[ \frac{1}{2} \log \left| \frac{\tan \frac{x}{2} + 1}{\tan \frac{x}{2} + 5} \right| + C \]