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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
MCQ
The value of $\int_{}^{} {\frac{{dx}}{{\sqrt {1 - x} }}} $ is
  • A
    $2\sqrt {1 - x} + c$
  • ✓
    $ - 2\sqrt {1 - x} + c$
  • C
    $ - {\sin ^{ - 1}}\sqrt x + c$
  • D
    ${\sin ^{ - 1}}\sqrt x + c$
✓

Answer

Correct option: B.
$ - 2\sqrt {1 - x} + c$
b
(b) We have, $\int_{}^{} {\frac{{dx}}{{\sqrt {1 - x} }}} $ or $I = \int_{}^{} {{{(1 - x)}^{ - 1/2}}dx} $
$I = \frac{{{{(1 - x)}^{\frac{{ - 1}}{2} + 1}}}}{{( - 1)\,\left( { - \frac{1}{2} + 1} \right)}} + c$ ==> $I = - 2\sqrt {1 - x} + c$.

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