$\int\sqrt{1+\sec x}dx$ is equal to:
Answer & explanation
Correct answer: option 1
$\int\sqrt{1+\sec x}dx=\int\frac{\sqrt{\sec^2x-1}}{\sqrt{\sec x-1}}dx=\int\frac{\tan x}{\sqrt{\sec x-1}}dx$
[Put $\sec x - 1=t^2⇒\sec x \tan x \,dx=2t\,dt$ $⇒\tan x\,dx=\frac{2t\,dt}{t^2+1}$]
$I=\int\frac{1}{t}(\frac{2t}{t^2+1})dt=2\tan^{-1}(t)+C=2\tan^{-1}(\sqrt{\sec x-1})+C$