Question

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

• A. $ln|\sec x(\sec x-\tan x)|+c$
• B. $ln|cosecx(\sec x+\tan x)|+c$
• C. $ln|\sec x(\sec x+\tan x)|+c$
• D. $ln|(\sec x+\tan x)|+c$

Answer 1

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

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

usng trigonometric identity ${\sec }^{2}x-{\tan }^{2}x=1$
$\int \{{\sec }^{2}x+{\tan }^{2}x+2\tan x\sec x{\}}^{\frac{1}{2}}dx$
$=\int (\sec x+\tan x)dx$
$=ln\sec x+ln(\sec x+\tan x)+c$
$=ln|\sec x(\sec x+\tan x)|+c$