$\displaystyle\int_0^{\pi/2} \log\sin x\,dx$ equals
A$\dfrac{\pi}{2}\log 2$
B$-\dfrac{\pi}{2}\log 2$
C$\dfrac{\pi}{4}\log 2$
D$-\pi\log 2$
Answer & Solution
Correct answer: B. $-\dfrac{\pi}{2}\log 2$
1. By $\mathbf{P_4}$, $I=\int_0^{\pi/2}\log\cos x\,dx$ too.
2. Adding: $2I=\int_0^{\pi/2}\log(\sin x\cos x)\,dx=\int_0^{\pi/2}\log\tfrac{\sin 2x}{2}\,dx$.
3. This gives $2I=\int_0^{\pi/2}\log\sin 2x\,dx-\tfrac{\pi}{2}\log 2$; the first integral reduces to $I$.
4. So $2I=I-\tfrac{\pi}{2}\log 2$, giving $I=-\dfrac{\pi}{2}\log 2$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.346_
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