The value of \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin x}{1+\cos x}\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\log_e 2\)
Let \(t = 1+\cos x\), so \(dt = -\sin x\,dx\).
When \(x = 0\), \(t = 2\); when \(x = \frac{\pi}{2}\), \(t = 1\). By substitution with new limits,
\( \displaystyle \int_{0}^{\pi/2}\frac{\sin x}{1+\cos x}\,dx = \int_{2}^{1}\frac{-dt}{t} \)
\( \displaystyle = \int_{1}^{2}\frac{dt}{t} \)
\( = \log_e 2 - \log_e 1 = \log_e 2\)
Why the other options are incorrect
Option (1) \(-\log_e 2\) — Incorrect. It keeps the limits 2 to 1 but drops the minus sign of \(dt = -\sin x\,dx\). The integrand is positive on \((0, \frac{\pi}{2})\), so the value is positive.
Option (2) 1 — Incorrect. The value is \(\log_e 2 \approx 0.693\), not 1.
Option (4) \(\displaystyle\frac{1}{2}\log_e 2\) — Incorrect. No factor \(\frac{1}{2}\) appears: \(\sin x\,dx\) is exactly \(-dt\).