Definite Integration
Indefinite Integration
Grade Class 12
Question:
The integral ∫ <span>sec<sup>2</sup> x</span> / <span>(sec x + tan x)<sup>9/2</sup></span> dx equals (for some arbitrary constant K)
- <span>1</span> / <span>(sec x + tan x)<sup>11/2</sup></span> { <span>1/11</span> - <span>1/7</span> (sec x + tan x)<sup>2</sup> } + K
- <span>1</span> / <span>(sec x + tan x)<sup>11/2</sup></span> { <span>1/11</span> - <span>1/7</span> (sec x + tan x)<sup>2</sup> } + K
- <span>1</span> / <span>(sec x + tan x)<sup>11/2</sup></span> { <span>1/11</span> + <span>1/7</span> (sec x + tan x)<sup>2</sup> } + K
<span>1</span> / <span>(sec x + tan x)<sup>11/2</sup></span> { <span>1/11</span> + <span>1/7</span> (sec x + tan x)<sup>2</sup> } + 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. Also, sec 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 = u sec x 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 = (u^2 + 1)/(2u). The integral becomes \int (sec x * sec x dx) / u^9/2 = \int (sec x * du/u) / u^9/2 = \int (u^2 + 1)/(2u) * (du/u) / u^9/2 = 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. Substituting u = sec x + tan x, we get - 1/(sec x + tan x)^11/2 { 1/11 + 1/7 (sec x + tan x)^2 } + K.
Correct Answer: C