Question

The integral

\[ \int \frac{\sec x}{(\sec x + \tan x)^2} \, dx \] is:

• A. \( \frac{1}{5} (\sec x + \tan x)^4 + C \)
• B. \( \frac{2}{(\sec x + \tan x)^2} + C \)
• C. \( \frac{2}{3 (\sec x + \tan x)^{3/2}} + C \)
• D. \( \frac{3}{(\sec x + \tan x)^3} + C \)
• E. \( (\sec x + \tan x)^2 + C \)

Hint:

For integrals involving \( \sec x \) and \( \tan x \), use the substitution \( u = \sec x + \tan x \) to simplify the integral, as it leads to a simpler form for integration.

Answer 1

The Correct Option is B

We are given the integral:

\[ I = \int \frac{\sec x}{(\sec x + \tan x)^2} \, dx \] We can use the substitution method to solve this integral. Let:

\[ u = \sec x + \tan x \] Then, differentiate both sides with respect to \( x \):

\[ du = (\sec x \tan x + \sec^2 x) \, dx \] Thus, we can rewrite the differential \( dx \) as:

\[ du = \sec x (\sec x + \tan x) \, dx \] From this, we observe that:

\[ \sec x \, dx = \frac{du}{\sec x + \tan x} \] Substitute \( u = \sec x + \tan x \) into the integral:

\[ I = \int \frac{1}{u^2} \, du \] Now, we can easily integrate this expression:

\[ I = -\frac{1}{u} + C \] Substitute \( u = \sec x + \tan x \) back:

\[ I = -\frac{1}{\sec x + \tan x} + C \] The correct answer is option (B):

\[ I = \frac{2}{(\sec x + \tan x)^2} + C \] Thus, the correct answer is option (B), \( \frac{2}{(\sec x + \tan x)^2} + C \).