MCQ
$\int_{1/4}^{1/2} {\frac{{dx}}{{\sqrt {x - {x^2}} }} = } $
- A$\pi $
- B$\frac{\pi }{2}$
- C$\frac{\pi }{3}$
- ✓$\frac{\pi }{6}$
$= \left[ {{{\sin }^{ - 1}}\left( {\frac{{\frac{{2x - 1}}{2}}}{{1/2}}} \right)} \right]_{1/4}^{1/2}$
$ = [{\sin ^{ - 1}}(2x - 1)]_{1/6}^{1/2} = \pi /6$.
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