AP EAMCET · Maths · Indefinite Integration
\(\int \frac{\sec ^2 x}{(\sec x+\tan x)^{5 / 2}} d x=\)
- A \(-\frac{(\sec x+\tan x)^{5 / 2}}{5}-\frac{(\sec x+\tan x)^{7 / 2}}{7}+c\)
- B \(-\frac{(\sec x-\tan x)^{5 / 2}}{5}-\frac{(\sec x-\tan x)^{7 / 2}}{7}+c\)
- C \(-\frac{(\sec x+\tan x)^{3 / 2}}{3}-\frac{(\sec x+\tan x)^{7 / 2}}{7}+c\)
- D \(-\frac{(\sec x-\tan x)^{3 / 2}}{3}-\frac{(\sec x-\tan x)^{7 / 2}}{7}+c\)
Answer & Solution
Correct Answer
(D) \(-\frac{(\sec x-\tan x)^{3 / 2}}{3}-\frac{(\sec x-\tan x)^{7 / 2}}{7}+c\)
Step-by-step Solution
Detailed explanation
Let \( u = \sec x + \tan x \). \( du = (\sec x \tan x + \sec^2 x) dx = \sec x (\sec x + \tan x) dx = u \sec x \ dx \). \( \sec x \ dx = \frac{du}{u} \). Since \( \sec^2 x - \tan^2 x = 1 \), \( (\sec x - \tan x)(\sec x + \tan x) = 1 \). Thus, \( \sec x - \tan x = u^{-1} \).…
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