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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
$\int_{}^{} {\frac{x}{{\sqrt {4 - {x^4}} }}dx} = $
  • A
    ${\cos ^{ - 1}}\frac{{{x^2}}}{2}$
  • B
    $\frac{1}{2}{\cos ^{ - 1}}\frac{{{x^2}}}{2}$
  • C
    ${\sin ^{ - 1}}\frac{{{x^2}}}{2}$
  • ✓
    $\frac{1}{2}{\sin ^{ - 1}}\frac{{{x^2}}}{2}$
✓

Answer

Correct option: D.
$\frac{1}{2}{\sin ^{ - 1}}\frac{{{x^2}}}{2}$
d
(d)$\int_{}^{} {\frac{x}{{\sqrt {4 - {x^4}} }}} \,dx = \int_{}^{} {\frac{x}{{\sqrt {{2^2} - {{({x^2})}^2}} }}} \,dx$
Putting ${x^2} = t \Rightarrow 2x\,dx = dt,$

we get the required integral $ = \frac{1}{2}{\sin ^{ - 1}}\frac{{{x^2}}}{2}$.

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