KCET · Maths · Indefinite Integration
\(\int \frac{\sec x}{\sec x+\tan x} d x\) is equal to
- A \(\tan x-\sec x+C\)
- B \(\log (1+\sec \mathrm{x})+\mathrm{C}\)
- 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
\[
\begin{aligned}
I &=\int \frac{\sec x}{\sec x+\tan x} d x \\
&=\int \frac{\sec x(\sec x-\tan x)}{\sec ^{2} x-\tan ^{2} x} d x \\
&=\int\left(\sec ^{2} x-\sec x \tan x\right) d x \\
&=\tan x-\sec x+C
\end{aligned}
\]
\[
\begin{aligned}
I &=\int \frac{\sec x}{\sec x+\tan x} d x \\
&=\int \frac{\sec x(\sec x-\tan x)}{\sec ^{2} x-\tan ^{2} x} d x \\
&=\int\left(\sec ^{2} x-\sec x \tan x\right) d x \\
&=\tan x-\sec x+C
\end{aligned}
\]
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