\int{\sec ^{2} x}/{(\sec x+\tan x)^{9 \/ 2}} d x equals

Correct Answer is : −1(secx+tanx)11∕2 {111+17(secx+tanx)}+K

- {1}/{(\sec x+\tan x)^{11 \/ 2}} \{ \frac{1}{11}- \frac{1}{7}(\sec x+\tan x)^{2} \}+K

{1}/{(\sec x+\tan x)^{11 \/ 2}} \{ \frac{1}{11}- \frac{1}{7}(\sec x+\tan x)^{2} \}+K

- {1}/{(\sec x+\tan x)^{11 \/ 2}} \{ \frac{1}{11}+ \frac{1}{7}(\sec x+\tan x) \}+K

Manipal Entrance Test 2015 Math Paper

Question : 26 of 70

Marks: +1, -0

∫

sec2x

(sec‌x+tan‌x)9∕2

dx equals

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