Indefinite Integration
Trigonometric Substitution
Grade 12
Question:
<p>\(\int \frac{\sec x}{(\sec x + \tan x)^{9/2}} dx\) equals (for some arbitrary constant K) [IIT - 2012]</p>
<p>(A) \(-\frac{2}{7}(\sec x + \tan x)^{-7/2} + K\)</p>
<p>(B) \(-\frac{1}{7}(\sec x + \tan x)^{-7/2} + K\)</p>
<p>(C) \(\frac{2}{7}(\sec x + \tan x)^{-7/2} + K\)</p>
<p>(D) None of these</p>
Step-by-Step Solution
Key Concept: Recognize the derivative relationship and use substitution to convert to a simple power function.
<p>Use substitution $u = \sec x + \tan x$. Then $du = (\sec x \tan x + \sec^2 x) dx = \sec x(\tan x + \sec x) dx = \sec x \cdot u \, dx$. The integral becomes $\int u^{-9/2} du = \frac{u^{-7/2}}{-7/2} = -\frac{2}{7}(\sec x + \tan x)^{-7/2} + K$.</p>
Correct Answer: A