Question

$\int \frac{\tan x}{\sec x+\tan x}dx$ is equal to

• A. $\sec x+\tan x-x+C$
• B. $\sec x-\tan x+x+C$
• C. $\sec x+\tan x+x+C$
• D. $-\sec x+\tan x+x+C$

Answer 1

Answer: (B)

$\sec x-\tan x+x+C$

Let $I=\int \frac{\tan x}{(\sec x+\tan x)}dx$

On multiplying numerator and denominator by $(\sec x-\tan x)$ we get

$I=\int \frac{\tan x(\sec x-\tan x)}{(\tan x+\sec x)(\sec x-\tan x)}dx$
$=\int \frac{\tan x(\sec x-\tan x)}{({\sec }^{2}x-{\tan }^{2}x)}dx=\int (\sec x\tan x-{\tan }^{2}x)dx$
$=\int \sec x\tan xdx-\int ({\sec }^{2}x-1)dx=\int \sec x\tan xdx-\int \left({\sec }^{2}x\right)dx+\int 1dx$
$\Rightarrow I=\sec x-\tan x+x+C$