Question

If $\int \sqrt{1+\sec x}dx=2(fog)(x)+C$, then

• A. $f(x)=\sec x-1$
• B. $f(x)=2{\left(\tan \right)}^{-1}x$
• C. $g(x)=\sqrt{\sec x-1}$
• 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 $x=t\Rightarrow \sec x\tan xdx=dt$ )
$=\int \frac{dt}{\mathrm{t}\sqrt{\mathrm{t}-1}}=2\int \frac{ydy}{y^2+1y}$ 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 $g(x)=\sqrt{\sec x-1}.$