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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
The integral $\int_0^\pi \sqrt{1+4 \sin ^2 \frac{x}{2}-4 \sin \frac{x}{2}} d x$ equals
  • A
    $4 \sqrt{3}-4$
  • ✓
    $4 \sqrt{3}-4-\frac{\pi}{3}$
  • C
    $\pi-4$
  • D
    $\frac{2 \pi}{3}-4-4 \sqrt{3}$
✓

Answer

Correct option: B.
$4 \sqrt{3}-4-\frac{\pi}{3}$
(B)
$\int_0^\pi \sqrt{1+4 \sin ^2 \frac{x}{2}-4 \sin \frac{x}{2}} d x$
$=\int_0^\pi\left|2 \sin \frac{x}{2}-1\right| d x \\ =\int_0^{\frac{\pi}{3}}\left|2 \sin \frac{x}{2}-1\right| d x+\int_{\frac{\pi}{3}}^\pi\left|2 \sin \frac{x}{2}-1\right| d x \\ =\int_0^{\frac{\pi}{3}}\left(1-2 \sin \frac{x}{2}\right) d x +\int_{\frac{\pi}{3}}^\pi\left(2 \sin \frac{x}{2} 1\right) d x$
$=\left[x+4 \cos \frac{x}{2}\right]_0^{\frac{\pi}{3}}+\left[-4 \cos \frac{x}{2}-x\right]_{\frac{\pi}{3}}^\pi \\ =\frac{\pi}{3}+4\left(\frac{\sqrt{3}}{2}-1\right)+\left[-4\left(0-\frac{\sqrt{3}}{2}\right)-\left(\pi-\frac{\pi}{3}\right)\right] \\ =4 \sqrt{3}-4-\frac{\pi}{3}$

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