MCQ
$\int \frac{\sec x}{\sec x-\tan x} d x \text { equals }$
  • A
    $\sec x-\tan x+c$
  • B
    $\sec x+\tan x+c$
  • $\tan x-\sec x+c$
  • D
    $-(\sec x+\tan x)+c$

Answer

Correct option: C.
$\tan x-\sec x+c$
$\text {Let } I=\int \frac{\sec x}{\sec x-\tan x} d x$
$=\int \frac{\sec x(\sec x+\tan x)}{(\sec x-\tan x)(\sec x+\tan x)} d x=\int\left(\frac{\sec ^2 x+\sec x \tan x}{\sec ^2 x-\tan ^2 x}\right) d x$
$=\int \sec ^2 x d x+\int \sec x \tan x d x \quad\left[\because \sec ^2 x-\tan ^2 x=1\right]$
$=\tan x+\sec x+c$

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