If \( \int\displaystyle\frac{1}{\tan x+\cot x}\,dx=k\cos 2x+C \), where C is an arbitrary constant, then the value of \( k \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → -1/4
\( \displaystyle \frac{1}{\tan x+\cot x}=\frac{\sin x\cos x}{\sin^2x+\cos^2x} \)
\( \displaystyle =\sin x\cos x \)
\( \displaystyle =\frac{\sin 2x}{2} \)
\( \displaystyle \int\frac{\sin 2x}{2}\,dx=-\frac{\cos 2x}{4}+C \)
So k = −1/4.
Why the other options are incorrect
Option (1) 1/2 — Incorrect. The value is −1/4, not 1/2.
Option (2) -1/2 — Incorrect. This misses the factor 1/2 that comes from integrating \( \sin 2x \).
Option (3) 1/4 — Incorrect. This misses the minus sign: \( \int\sin 2x\,dx=-\frac{\cos 2x}{2} \).