Question

Find the value of: \(\displaystyle\int_0^{\pi/2}\log\sin x\,dx\).

Hint:

Use the King's rule (property P4) to relate I to log cos x, add, use sin2x, and self-reference the same integral.

Answer 1

Step 1: Let \(I=\displaystyle\int_0^{\pi/2}\log\sin x\,dx\) and use the property \(\int_0^a f(x)dx=\int_0^a f(a-x)dx\):
\[ I=\int_0^{\pi/2}\log\sin\left(\dfrac\pi2-x\right)dx=\int_0^{\pi/2}\log\cos x\,dx \]

Step 2: Add the two expressions for \(I\):
\[ 2I=\int_0^{\pi/2}\big(\log\sin x+\log\cos x\big)dx=\int_0^{\pi/2}\log(\sin x\cos x)\,dx=\int_0^{\pi/2}\log\left(\dfrac{\sin2x}2\right)dx \]
\[ 2I=\int_0^{\pi/2}\log\sin2x\,dx-\int_0^{\pi/2}\log2\,dx \]

Step 3: Evaluate \(\int_0^{\pi/2}\log\sin2x\,dx\) via the substitution \(t=2x\):
\[ \int_0^{\pi/2}\log\sin2x\,dx=\dfrac12\int_0^\pi\log\sin t\,dt=\dfrac12\cdot2\int_0^{\pi/2}\log\sin t\,dt=I \]
(using symmetry of \(\sin t\) about \(t=\pi/2\) on \([0,\pi]\)).

Step 4: Substitute back and solve for \(I\):
\[ 2I=I-\dfrac\pi2\log2 \implies I=-\dfrac\pi2\log2 \]

Final Answer:
\[ \boxed{I=-\dfrac\pi2\ln2} \]