MCQ
$\mathop \smallint \limits_0^\pi \sqrt {1 + 4{{\sin }^2}\frac{x}{2} - 4\sin \frac{x}{2}} \;dx = $
- A$4\sqrt 3 - 4$
- ✓$\;4\sqrt 3 - 4 - \frac{\pi }{3}$
- C$\pi - 4\;$
- D$\frac{{2\pi }}{3} - 4\sqrt 3 - 4$
✓
Answer
Correct option: B.
$\;4\sqrt 3 - 4 - \frac{\pi }{3}$
b
$\int\limits_0^\pi {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx$
$\int\limits_0^\pi {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx$
$ = \int\limits_0^{\frac{\pi }{3}} {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx - \int\limits_{\frac{\pi }{3}}^\pi {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx$
$=\left(x+4 \cos \frac{x}{2}\right)_{0}^{\frac{\pi}{3}}-\left(x+4 \cos \frac{x}{2}\right)_{\frac{\pi}{3}}^{\pi}$
$=\frac{\pi}{3}+4 \cos \frac{\pi}{6}-0-4-\left(\pi+4 \cos \frac{\pi}{2}-\frac{\pi}{3}-4 \cos \frac{\pi}{6}\right)$
$=-\frac{\pi}{3}+4 \sqrt{3}-4$
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