Definite Integration
Integration by substitution
Grade Class 12
Question:
The integral ∫ sec^2 x / (sec x + tan x)^9/2 dx equals (for some arbitrary constant K)
- \frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7}(\sec x + \tan x)^2 \right\} + K
- \frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} - \frac{1}{7}(\sec x + \tan x)^2 \right\} + K
- \frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x)^2 \right\} + K
- \frac{1}{(\sec x + \tan x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x + \tan x)^2 \right\} + K
Step-by-Step Solution
Key Concept: Use substitution u = sec x + tan x, then du = sec x(sec x + tan x) dx. Also note sec x = (u + 1/u)/2 and tan x = (u - 1/u)/2.
Let u = sec x + tan x. Then du = (sec x tan x + sec^2 x) dx = sec x(tan x + sec x) dx = sec x * u dx. So sec x dx = du/u. Since sec x - tan x = 1/u, we have 2 sec x = u + 1/u, so sec x = (u^2 + 1)/(2u). Thus dx = (2u / (u^2 + 1)) * (du/u) = 2/(u^2 + 1) du. The integral becomes \int (sec x / u^9/2) * (sec x dx) = \int ((u^2+1)/2u * 1/u^9/2) * (du/u) ...
Correct Answer: D