\(\int\displaystyle\frac{\sin x}{\sin 2x}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{1}{2}\log|\sec x+\tan x|+C\)
Using the double-angle formula \(\sin 2x = 2\sin x\cos x\):
\( \displaystyle \frac{\sin x}{\sin 2x} = \frac{\sin x}{2\sin x\cos x} = \frac{1}{2}\sec x \)
With the standard integral \(\int\sec x\,dx = \log|\sec x+\tan x|\):
\( \displaystyle \int\frac{1}{2}\sec x\,dx = \frac{1}{2}\log|\sec x+\tan x|+C \)
Why the other options are incorrect
Option (1) \(\frac{1}{2}\log|\sec x-\tan x|+C\) — Incorrect. Since \((\sec x-\tan x)(\sec x+\tan x) = 1\), this is the negative of the answer.
Option (2) \(\frac{1}{2}\log|\sin x+\cos x|+C\) — Incorrect. Its derivative is \(\frac{\cos x-\sin x}{2(\sin x+\cos x)}\), which is not \(\frac{1}{2}\sec x\).
Option (3) \(\log|\sec x+\tan x|+C\) — Incorrect. It drops the factor \(\frac{1}{2}\) from \(\sin 2x = 2\sin x\cos x\).