Definite Integration
Integration by substitution
Grade Class 12
Question:
The integral ∫ <sup>sec²x</sup> / <sub>(sec x + tan x)<sup>9/2</sup></sub> dx equals (for some arbitrary constant K)
- <sup>1</sup>/<sub>(sec x + tan x)<sup>11/2</sup></sub> {<sup>1</sup>/<sub>11</sub> - <sup>1</sup>/<sub>7</sub>(sec x + tan x)<sup>2</sup>} + K
- <sup>1</sup>/<sub>(sec x + tan x)<sup>11/2</sup></sub> {<sup>1</sup>/<sub>11</sub> - <sup>1</sup>/<sub>7</sub>(sec x + tan x)<sup>2</sup>} + K
- <sup>1</sup>/<sub>(sec x + tan x)<sup>11/2</sup></sub> {<sup>1</sup>/<sub>11</sub> + <sup>1</sup>/<sub>7</sub>(sec x + tan x)<sup>2</sup>} + K
- <sup>1</sup>/<sub>(sec x + tan x)<sup>11/2</sup></sub> {<sup>1</sup>/<sub>11</sub> + <sup>1</sup>/<sub>7</sub>(sec x + tan x)<sup>2</sup>} + K
Step-by-Step Solution
Key Concept: Let 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. Also, sec x - tan x = 1/u. Adding gives 2 sec x = u + 1/u, so sec x = 1/2(u + 1/u). The integral becomes \int (1/2(u + 1/u)) / u^9/2 * (du/u) = 1/2 \int (u + 1/u) / u^11/2 du = 1/2 \int (u^-9/2 + u^-13/2) du.
Let u = sec x + tan x. Then du = (sec x tan x + sec^2 x) dx = sec x (sec x + tan x) dx = sec x * u dx. Thus, 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 = 1/2(u + 1/u). The integral becomes \int (sec x * sec x dx) / u^9/2 = \int (1/2(u + 1/u) * du/u) / u^9/2 = 1/2 \int (u + 1/u) / u^11/2 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. Substituting u = sec x + tan x, we get the result.
Correct Answer: C