Question

Evaluate: \(\displaystyle\int_0^{\pi/2}\log\sin x\,dx\).

Hint:

Use \(I=\int_0^{\pi/2}\log\cos x\,dx\) (by \(x\to\pi/2-x\)), add to get \(2I\) via \(\log(\sin x\cos x)\), and resolve the \(\log\sin2x\) piece by substitution.

Answer 1

Step 1: Understanding the Concept:
Use the King's-rule property \(\int_0^a f(x)dx=\int_0^a f(a-x)dx\), which for this integral swaps \(\sin x\) with \(\cos x\); combining the two forms and using the double-angle identity for \(\sin2x\) collapses the integral onto itself, allowing us to solve for it algebraically.

Step 2: Applying x -> pi/2 - x:
Let \(I=\displaystyle\int_0^{\pi/2}\log\sin x\,dx\). Using \(x\to\frac{\pi}{2}-x\): \(I=\displaystyle\int_0^{\pi/2}\log\sin\left(\frac{\pi}{2}-x\right)dx=\int_0^{\pi/2}\log\cos x\,dx\).

Step 3: Adding the two forms:
\(2I=\displaystyle\int_0^{\pi/2}(\log\sin x+\log\cos x)\,dx=\int_0^{\pi/2}\log(\sin x\cos x)\,dx\).

Step 4: Using the double angle identity:
\(\sin x\cos x=\dfrac{\sin2x}{2}\), so \(2I=\displaystyle\int_0^{\pi/2}\log\left(\dfrac{\sin2x}{2}\right)dx=\int_0^{\pi/2}\log\sin2x\,dx-\int_0^{\pi/2}\log2\,dx\).

Step 5: Evaluating each piece:
For the first piece, substitute \(t=2x\), \(dt=2dx\): \(\displaystyle\int_0^{\pi/2}\log\sin2x\,dx=\frac12\int_0^{\pi}\log\sin t\,dt=\frac12\cdot2\int_0^{\pi/2}\log\sin t\,dt=I\) (using the symmetry of \(\sin t\) about \(t=\pi/2\) over \([0,\pi]\)). The second piece is \(\dfrac{\pi}{2}\log2\).

Step 6: Solving for I:
So \(2I=I-\dfrac{\pi}{2}\log2\ \Rightarrow\ I=-\dfrac{\pi}{2}\log2\).

Final Answer:
\(\displaystyle\int_0^{\pi/2}\log\sin x\,dx=\boxed{-\dfrac{\pi}{2}\log2}\).