Question

Prove that \(\displaystyle\int_0^{\pi/2}\log\sin x\,dx=-\dfrac{\pi}{2}\log2\).

Hint:

Use the a→(a−x) reflection trick to relate ∫log sinx to ∫log cosx, combine via sin2x=2sinxcosx.

Answer 1

Step 1: Defining I and applying the reflection property:
Let \(I=\displaystyle\int_0^{\pi/2}\log\sin x\,dx\). Using \(\displaystyle\int_0^a f(x)dx=\int_0^a f(a-x)dx\) with \(a=\pi/2\): \(I=\displaystyle\int_0^{\pi/2}\log\sin\left(\dfrac\pi2-x\right)dx=\int_0^{\pi/2}\log\cos x\,dx\).

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

Step 3: Using the double-angle identity:
\(\sin x\cos x=\dfrac{\sin2x}{2}\), so \(2I=\displaystyle\int_0^{\pi/2}\left[\log\sin2x-\log2\right]dx=\int_0^{\pi/2}\log\sin2x\,dx-\dfrac\pi2\log2\).

Step 4: Evaluating the remaining integral by substitution:
Let \(u=2x\), \(du=2dx\); as \(x:0\to\pi/2\), \(u:0\to\pi\): \(\displaystyle\int_0^{\pi/2}\log\sin2x\,dx=\dfrac12\int_0^\pi\log\sin u\,du\). Since \(\sin(\pi-u)=\sin u\), the graph of \(\log\sin u\) is symmetric about \(u=\pi/2\), so \(\displaystyle\int_0^\pi\log\sin u\,du=2\int_0^{\pi/2}\log\sin u\,du=2I\). Hence \(\displaystyle\int_0^{\pi/2}\log\sin2x\,dx=\dfrac12(2I)=I\).

Step 5: Solving for I:
So \(2I=I-\dfrac\pi2\log2\Rightarrow I=-\dfrac\pi2\log2\).

Final Answer:
\[ \boxed{\displaystyle\int_0^{\pi/2}\log\sin x\,dx=-\dfrac\pi2\log2} \]