Question

If $\int \sqrt{1+\sec x} d x=2(f o g)(x)+C$, then

• A. $f\left(\right.x)=\sec x-1$
• B. $f\left(\right.x\left.\right)=2\left(\tan\right)^{- 1}x$
• C. $g\left(\right.x\left.\right)=\sqrt{\sec x - 1}$
• D. None of these

Answer 1

Answer: (C)

$\int \sqrt{1+\sec x } dx =\int \frac{\sec x \cdot \tan x }{\sec x \cdot \sqrt{\sec x -1}} dx =\int$
$\frac{\tan x }{\sec x -1} dx$
(Put sec $x = t \Rightarrow \sec x \tan x dx = dt$ )
$=\int \frac{ dt }{\operatorname{t\sqrt {t-1}}}=2 \int \frac{ ydy }{ y ^{2}+1 y }$ where $y ^{2}= t -1$
$=2 \tan ^{-1} y + C =2 \tan ^{-1} \sqrt{\sec x -1}+ C$
$\therefore f ( x )=\tan ^{-1} x$ and $g ( x )=\sqrt{\sec x -1} .$