COMEDK · Maths · 28. Indefinite Integration
\(\int \frac{\sec x}{\sec x+\tan x} d x=\)
- A \(\tan x-\sec x+C\)
- B \(\log (1+\sin x)+C\)
- C \(\sec x+\tan x+C\)
- D \(\log \sin x+\log \cos x+C\)
Answer & Solution
Correct Answer
(A) \(\tan x-\sec x+C\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{\sec x}{\sec x+\tan x} d x\) \[ =\int \frac{1}{1+\sin x} d x=\int \frac{1-\sin x}{1-\sin ^{2} x} d x \] \(=\int \frac{1-\sin x}{\cos ^{2} x} d x\) \(=\int\left(\sec ^{2} x-\sec x \tan x\right) d x\) \(=\tan x-\sec x+\bar{C}\)
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