Question

$\int \{1+2\tan x(\tan x+\sec x){\}}^{\frac{1}{2}}dx=$

• A. $log(\sec x+\tan x)+c$
• B. $log(\sec x+\tan x{)}^{\frac{1}{2}}+c$
• C. $log\sec x(\sec x+\tan x)+c$
• D. None of these

Answer 1

The correct option is C $log\sec x(\sec x+\tan x)+c$

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

$=\int \sqrt{1+2{\tan }^{2}x+2\tan x\sec x}dx$

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

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

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

$=\int (\sec x+\tan x)dx$

$=\int \sec xdx+\int \tan xdx$

$=log(\sec x+\tan x)+log\left(\sec x\right)+c$

$=log\left(\sec x\right(\sec x+\tan x)+c$

$\therefore \int \sqrt{1+2\tan x(\tan x+\sec x)}dx=log\left(\sec x\right(\sec x+\tan x\left)\right)+c$.