alleryStart learning →$\displaystyle\int_0^{\pi/2} \log(\sin x)\, dx = ?$ (classic)
- $-(\pi/2)\ln 2$
- $\ln 2$
- $0$
- $-\pi\ln 2$
Show answer and solution
Correct answer: A. $-(\pi/2)\ln 2$
$I = \int_0^{\pi/2} \log\sin x\, dx$.
By King's, $I = \int_0^{\pi/2} \log\cos x\, dx$.
So $2I = \int_0^{\pi/2} \log(\sin x \cos x)\, dx = \int_0^{\pi/2} [\log(\sin 2x) - \log 2]\, dx$.
Sub $u = 2x$ in first: $= \tfrac{1}{2}\int_0^\pi \log\sin u\, du - (\pi/2)\log 2 = \tfrac{1}{2}(2I) - (\pi/2)\log 2 = I - (\pi/2)\log 2$.
**So $I = -(\pi/2)\log 2$.**