Question

$\int \frac{\sec x+\tan x}{\sec x-\tan x}dx=$

• A. $2(\tan x+\sec x)-x+c$
• B. $\tan x-\sec x+x+c$
• C. $2(\tan x+\sec x)+c$
• D. $2(\tan x+\sec x)+x+c$

Answer 1

Answer: (A)

$2(\tan x+\sec x)-x+c$

$\because {\sec }^{2}x-{\tan }^{2}x=1$

Hence, $\sec x+\tan x=\frac{1}{\sec x-\tan x}$

So, $\frac{\sec x+\tan x}{\sec x-\tan x}={(\sec x+\tan x)}^2$

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

Since the integral sign distributes across summation,

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

$=\int 2{\sec }^{2}xdx-\int dx+\int 2\sec x\tan xdx$ ..... $\because {\sec }^{2}x-1={\tan }^{2}x$

$=2\tan x+2\sec x-x+c$

$=2(\tan x+\sec x)-x+c$