Question

If $\int \frac{\tan x}{\sec x+\tan x}dx=I+x+c$ where $c$ is an arbitary constant, then the value of $I$ is

• A. $\sec x-\tan x$
• B. $\sec x+\tan x$
• C. $2\log \left(\sec \frac{x}{2}\right)$
• D. $\frac{1}{2}\log (\sec x+\tan x)$

Answer 1

The correct option is A $\sec x-\tan x$

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