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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\int_{\,0}^{\,1} {\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} \right]\,dx} $ is equal to
  • A
    $0$
  • B
    $\pi $
  • ✓
    $\pi /2$
  • D
    $\pi /4$
✓

Answer

Correct option: C.
$\pi /2$
c
(c)$I = \left[ {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)} \right]_0^1 = {\sin ^{ - 1}}(1) - {\sin ^{ - 1}}(0)$$ = \frac{\pi }{2}$.

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