Definite Integration
Integration of trigonometric functions
Grade Class 12
Question:
The integral ∫ sec² x / (sec x + tan x)^9/2 dx equals (for some arbitrary constant K)
- 1 / (sec x + tan x)^11/2 { 1/11 - 1/7 (sec x + tan x)² } + K
- 1 / (sec x + tan x)^11/2 { 1/11 - 1/7 (sec x + tan x)² } + K
- 1 / (sec x + tan x)^11/2 { 1/11 + 1/7 (sec x + tan x)² } + K
- 1 / (sec x + tan x)^11/2 { 1/11 + 1/7 (sec x + tan x)² } + K
Step-by-Step Solution
Key Concept: Use the substitution u = sec x + tan x, then du = (sec x tan x + sec^2 x) dx = sec x (tan x + sec x) dx = u sec x dx. Thus sec x dx = du/u. Also sec x - tan x = 1/u, so 2 sec x = u + 1/u, sec x = (u^2 + 1)/2u.
Let u = sec x + tan x. Then du = sec x (sec x + tan x) dx = sec x * u dx. So sec x dx = du/u. Also, sec x - tan x = 1/u. Adding the two equations, 2 sec x = u + 1/u, so sec x = (u^2 + 1)/2u. The integral becomes \int (sec x * sec x dx) / u^9/2 = \int ((u^2 + 1)/2u * du/u) / u^9/2 = 1/2 \int (u^2 + 1) / u^(9/2 + 2) du = 1/2 \int (u^2 + 1) / u^13/2 du = 1/2 \int (u^-9/2 + u^-13/2) du = 1/2 [ u^-7/2 / (-7/2) + u^-11/2 / (-11/2) ] + K = -1/7 u^-7/2 - 1/11 u^-11/2 + K = -1/u^11/2 [ 1/11 + 1/7 u^2 ] + K.
Correct Answer: D