Question

Evaluate the integral:  

\[ \int \frac{\sec^2 x}{(\sec x + \tan x)^{9/2}} \, dx \]

• A. \(-\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7}(\sec x + \tan x)^2 \right\} + K\)
• B. \(\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7}(\sec x + \tan x)^2 \right\} + K\)
• C. \(-\frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x) \right\} + K\)
• D. None of the above

Hint:

The substitution \(t = \sec x + \tan x\) is very useful for integrals involving \(\sec x\).

Answer 1

The Correct Option is A

Step 1: Concept:
Let \[ t = \sec x + \tan x \] Then, \[ \frac{1}{t} = \sec x - \tan x \] and, \[ \sec x = \frac{t^2 + 1}{2t} \]
Step 2: Detailed Explanation:
Differentiate: \[ dt = \sec x(\sec x + \tan x)\,dx = \sec x \cdot t \, dx \Rightarrow dx = \frac{dt}{t \sec x} \]
Substitute in the integral: \[ I = \int \frac{\sec^2 x}{t^{9/2}} \cdot \frac{dt}{t \sec x} = \int \frac{\sec x}{t^{11/2}} \, dt \] Now substitute \( \sec x = \frac{t^2 + 1}{2t} \): \[ I = \int \frac{t^2 + 1}{2t} \cdot \frac{1}{t^{11/2}} \, dt = \frac{1}{2} \int \left(t^{-13/2} + t^{-15/2}\right) dt \]
Integrate: \[ I = \frac{1}{2} \left[ \frac{t^{-11/2}}{-11/2} + \frac{t^{-13/2}}{-13/2} \right] + C \] \[ I = -\frac{1}{11} t^{-11/2} - \frac{1}{13} t^{-13/2} + C \]
Substitute back \( t = \sec x + \tan x \): \[ I = -\frac{1}{11(\sec x + \tan x)^{11/2}} - \frac{1}{13(\sec x + \tan x)^{13/2}} + C \]
This can be rearranged as: \[ I = -\frac{1}{(\sec x + \tan x)^{11/2}} \left[ \frac{1}{11} + \frac{1}{13(\sec x + \tan x)} \right] + C \]
Step 3: Final Answer:
\[ \boxed{ -\frac{1}{11(\sec x + \tan x)^{11/2}} - \frac{1}{13(\sec x + \tan x)^{13/2}} + C } \]