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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} } \,dx$ equals
  • ✓
    $\left( {\frac{\pi }{2} - 1} \right)$
  • B
    $\left( {\frac{\pi }{2} + 1} \right)$
  • C
    $\frac{\pi }{2}$
  • D
    $(\pi + 1)$
✓

Answer

Correct option: A.
$\left( {\frac{\pi }{2} - 1} \right)$
a
(a) $I = \int_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} dx = \int_0^1 {\sqrt {\frac{{1 - x}}{{1 + x}}} } .\frac{{\sqrt {1 - x} }}{{\sqrt {1 - x} }}dx} $

$ = \int_0^1 {\frac{{1 - x}}{{\sqrt {1 - {x^2}} }}dx = \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} - \int_0^1 {\frac{x}{{\sqrt {1 - {x^2}} }}\,} dx} $

$I = [{\sin ^{ - 1}}x]_0^1 + [\sqrt {1 - {x^2}} ]_0^1 = \frac{\pi }{2} - 1$.

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