Question

Evaluate the following integral: $ \int \frac{\sec x}{(\sec x + \tan x)^2} \, dx $

• A. \( - \frac{1}{\sec x + \tan x} \)
• B. \( \frac{1}{\sec x + \tan x} \)
• C. \( - \ln | \sec x + \tan x | \)
• D. \( \ln | \sec x + \tan x | \)

Hint:

Use substitution for integrals involving sums of trigonometric functions like \( \sec x + \tan x \) to simplify the expression.

Answer 1

The Correct Option is A

We are asked to evaluate the following integral: \[ \int \frac{\sec x}{(\sec x + \tan x)^2} \, dx \] We will use the substitution method. Let: \[ u = \sec x + \tan x \] Then, the derivative of \( u \) is: \[ du = (\sec x \tan x + \sec^2 x) \, dx \] Now, rewrite the integral in terms of \( u \). Notice that: \[ \frac{\sec x}{u^2} = \frac{du}{u^2} \] Now integrate: \[ \int \frac{1}{u^2} \, du = -\frac{1}{u} \] Substitute back \( u = \sec x + \tan x \): \[ -\frac{1}{\sec x + \tan x} \]
Thus, the correct answer is \( -\frac{1}{\sec x + \tan x} \).