\(\int\displaystyle\frac{\cos x-\cos 2x}{1-\cos x}\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(x+2\sin x+C\)
Using \(\cos 2x = 2\cos^2x-1\), the numerator is a quadratic in \(\cos x\) that factorises:
\(\cos x-\cos 2x = 1+\cos x-2\cos^2x\)
\(= (1-\cos x)(1+2\cos x)\)
The factor \(1-\cos x\) cancels:
\( \displaystyle \frac{\cos x-\cos 2x}{1-\cos x} = 1+2\cos x \)
Integrating:
\( \displaystyle \int(1+2\cos x)\,dx = x+2\sin x+C \)