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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_0^\pi x \log \sin x d x=$
  • A
    $\frac{\pi}{2} \log \frac{1}{2}$
  • ✓
    $\frac{\pi^2}{2} \log \frac{1}{2}$
  • C
    $\pi \log \frac{1}{2}$
  • D
    $\pi^2 \log \frac{1}{2}$
✓

Answer

Correct option: B.
$\frac{\pi^2}{2} \log \frac{1}{2}$
(B)
Let $I =\int_0^\pi x \log \sin x d x$ ...(i)
$\therefore \quad I=\int_0^\pi(\pi-x) \log \sin x d x$ ...(ii)
$\ldots \left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \log \sin x d x=2 \pi \int_0^{\frac{\pi}{2}} \log \sin x d x$
$\int_0^{\pi / 2} \log (\sin x) d x=\int_0^{\pi / 2} \log (\cos x) d x=-\frac{\pi}{2} \log 2$
$\therefore \quad 2 I =2 \pi\left(-\frac{\pi}{2} \log 2\right)$
$\Rightarrow I=\pi\left(\frac{\pi}{2} \log \frac{1}{2}\right)=\frac{\pi^2}{2} \log \frac{1}{2}$

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