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