∫1+secx dx=∫sec2x−1secx−1 dx secx−1=t2⇒secxtanx=2tdt ∫2tt(t2+1) =2tan−1t+C =2tan−1(secx−1)+C

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$\int \sqrt{1 + \sec x} d x =$

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$- 2 C o s^{- 1} (\sqrt{2} \cos \frac{x}{2}) + c$

$2 T a n h^{- 1} (\sqrt{\sec x - 1}) + c$

$- 2 C o s h^{- 1} (\sqrt{2} \cos \frac{x}{2}) + c$

$2 T a n^{- 1} (\sqrt{\sec x - 1}) + c$

answer is A.

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Detailed Solution

$\int \sqrt{1 + \sec x} d x = \int \frac{\sqrt{\sec^{2} x - 1}}{\sqrt{\sec x - 1}} d x$

$\sec x - 1 = t^{2} \Rightarrow \sec x \tan x = 2 t d t$

$\int \frac{2 t}{t (t^{2} + 1)} = 2 \tan^{- 1} t + C$

$= 2 \tan^{- 1} (\sqrt{\sec x - 1}) + C$

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