Question

Evaluate the integral: \[ \int \frac{\sec^2 x}{(\sec x+\tan x)^{5/2}}dx \]

• A. \( -\frac{(\sec x+\tan x)^{5/2}}{5}-\frac{(\sec x+\tan x)^{7/2}}{7}+c \)
• B. \( -\frac{(\sec x-\tan x)^{5/2}}{5}-\frac{(\sec x-\tan x)^{7/2}}{7}+c \)
• C. \( -\frac{(\sec x+\tan x)^{3/2}}{3}-\frac{(\sec x+\tan x)^{7/2}}{7}+c \)
• D. \( -\frac{(\sec x-\tan x)^{3/2}}{3}-\frac{(\sec x-\tan x)^{7/2}}{7}+c \)

Hint:

For integrals involving \( \sec x \) and \( \tan x \), substitution \( u = \sec x + \tan x \) simplifies differentiation.

Answer 1

The Correct Option is C

Using substitution: \[ u = \sec x + \tan x \] Differentiating: \[ du = (\sec x \tan x + \sec^2 x)dx = \sec^2 x dx \] Rewriting the integral: \[ \int \frac{du}{u^{5/2}} \] Applying the power rule: \[ \int u^{-5/2} du = \frac{u^{-3/2}}{-3/2} = -\frac{2}{3} u^{-3/2} \] Another integral term: \[ \int u^{-7/2} du = \frac{u^{-5/2}}{-5/2} = -\frac{2}{5} u^{-5/2} \] Rewriting: \[ -\frac{(\sec x+\tan x)^{3/2}}{3} - \frac{(\sec x+\tan x)^{7/2}}{7} + c \] Thus, the correct answer is: \[ -\frac{(\sec x+\tan x)^{3/2}}{3}-\frac{(\sec x+\tan x)^{7/2}}{7}+c \]