\(\int_{0}^{\frac{\pi}{2}}\displaystyle\log_e(\sin x)\,dx\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{\pi}{2}\log_e 2\)
Let \(I = \int_{0}^{\pi/2}\log_e(\sin x)\,dx\). Replacing x by \(\frac{\pi}{2} - x\),
\( \displaystyle I = \int_{0}^{\pi/2}\log_e(\cos x)\,dx \)
Adding the two forms, and using \(\sin x\cos x = \frac{1}{2}\sin 2x\):
\( \displaystyle 2I = \int_{0}^{\pi/2}\log_e(\sin x\cos x)\,dx \)
\( \displaystyle = \int_{0}^{\pi/2}\log_e(\sin 2x)\,dx - \frac{\pi}{2}\log_e 2 \qquad (1) \)
In the first integral put \(t = 2x\), \(dx = \frac{dt}{2}\), limits 0 to \(\pi\):
\( \displaystyle \int_{0}^{\pi/2}\log_e(\sin 2x)\,dx = \frac{1}{2}\int_{0}^{\pi}\log_e(\sin t)\,dt \)
Since \(\sin(\pi - t) = \sin t\), the integral over \([0, \pi]\) is twice the integral over \([0, \frac{\pi}{2}]\):
\( \displaystyle \frac{1}{2}\int_{0}^{\pi}\log_e(\sin t)\,dt = \frac{1}{2}\cdot 2I = I \)
Putting this in (1):
\( \displaystyle 2I = I - \frac{\pi}{2}\log_e 2 \)
\( \displaystyle I = -\frac{\pi}{2}\log_e 2 \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{\pi}{2}\log_e 2\) — Incorrect. \(0 \lt \sin x \lt 1\) on \((0, \frac{\pi}{2})\), so \(\log_e(\sin x) \lt 0\) and the integral must be negative.
Option (3) \(-\pi\log_e 2\) — Incorrect. This is the value of the integral over \([0, \pi]\), which is twice I.
Option (4) 0 — Incorrect. The integrand is negative throughout \((0, \frac{\pi}{2})\), so the integral cannot be 0.