_ 1/4 ^ 1/2 dx x - x^2 = - Vidyadip

MCQ

$\int_{1/4}^{1/2} {\frac{{dx}}{{\sqrt {x - {x^2}} }} = } $

A
$\pi $
B
$\frac{\pi }{2}$
C
$\frac{\pi }{3}$
✓
$\frac{\pi }{6}$

✓

Answer

Correct option: D.

$\frac{\pi }{6}$

d
(d) $\int_{1/4}^{1/2} {\frac{{dx}}{{\sqrt {x - {x^2}} }} = \int_{1/4}^{1/2} {\frac{{dx}}{{\sqrt {{{\left( {\frac{1}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}} }}} } $

$= \left[ {{{\sin }^{ - 1}}\left( {\frac{{\frac{{2x - 1}}{2}}}{{1/2}}} \right)} \right]_{1/4}^{1/2}$

$ = [{\sin ^{ - 1}}(2x - 1)]_{1/6}^{1/2} = \pi /6$.

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