Q119 / 180 · COMEDK 2019 · Mathematics

$\int \frac{\sec x}{\sec x+\tan x} d x=$

  • Correct answer
    A
    $\tan x-\sec x+C$
  • B
    $\log (1+\sin x)+C$
  • C
    $\sec x+\tan x+C$
  • D
    $\log \sin x+\log \cos x+C$

Solution

Let $I=\int \frac{\sec x}{\sec x+\tan x} d x$ $$ =\int \frac{1}{1+\sin x} d x=\int \frac{1-\sin x}{1-\sin ^{2} x} d x $$ $=\int \frac{1-\sin x}{\cos ^{2} x} d x$ $=\int\left(\sec ^{2} x-\sec x \tan x\right) d x$ $=\tan x-\sec x+\bar{C}$

Hence, the correct answer is option: 'A'.

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