Question

Integration of $\sqrt{1+2\tan x(\sec x+\tan x)}$ w. r. t. $x$ is

• A. $\ln |\sec x|-\ln |\sec x-\tan x|+c$
• B. $\ln |\sec x|+\ln |\sec x-\tan x|+c$
• C. $2\ln |\sec \frac{x}{2}+\tan \frac{x}{2}|+c$
• D. $\ln |1+\tan x(\sec x+\tan x)|+c$

Answer 1

Answer: (A)

$\ln |\sec x|-\ln |\sec x-\tan x|+c$

Let $I=\int \sqrt{1+2\tan x(\sec x+\tan x)}dx$

$\because \sqrt{1+2\tan x\sec x+2{\tan }^{2}x}$

$=\sqrt{{\sec }^{2}x-ta{n}^{2}x+2\tan x\sec x+2{\tan }^{2}x}$

$=\sqrt{{\sec }^{2}x+2\tan x\sec x+{\tan }^{2}x}$

$=\sqrt{{(\sec x+\tan x)}^2}=|\sec x+\tan x|$

$\therefore I=\int |\sec x+\tan x|dx$

$=\int |\sec x|dx+\int |\tan x|dx$

$=\ln |\sec x+\tan x|+\ln \left|\sec x\right|+c$

$=\ln \frac{1}{|\sec x-\tan x|}+\ln \left|\sec x\right|+c$ $[{\sec }^{2}x-{\tan }^{2}x=1\Rightarrow (secx+tanx)=\frac{1}{secx-tanx}]$

$=-\ln |\sec x-\tan x|+\ln \left|\sec x\right|+c$