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7.1 inderfinite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.1 inderfinite integral1 Mark
MCQ
If $\int {\frac{{\sqrt {1 - {x^2}} }}{{{x^4}}}} dx\, = \,A\,(x)\,{(\sqrt {1 - {x^2}} )^m}\, + \,C,$ for a suitable chosen integer $m$ and a function $A(x),$ where $C$ is a constant of integration, then $(A(x))^m$ equals
  • ✓
    $\frac{{ - 1}}{{27\,{x^9}}}$
  • B
    $\frac{{ - 1}}{{3\,{x^3}}}$
  • C
    $\frac{{  1}}{{27\,{x^6}}}$
  • D
    $\frac{{  1}}{{9\,{x^4}}}$
✓

Answer

Correct option: A.
$\frac{{ - 1}}{{27\,{x^9}}}$
a
$\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=\int \frac{x \sqrt{\frac{1}{x^{2}}-1}}{x^{4}} d x=\int \frac{1}{x^{3}} \sqrt{\frac{1}{x^{2}}-1} d x$

${\rm{ Let }}\frac{1}{{{x^2}}} - 1 = t$

$ - \frac{2}{{{x^3}}}{\rm{d}}x = {\rm{dt}}$

${\text { Integration becomes }-\frac{1}{2} \int \sqrt{t} d t=-\frac{1}{2} \frac{(t)^{3 / 2}}{\frac{3}{2}}}$

${=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{3 / 2}}$

${=-\frac{1}{3 x^{3}}\left(1-x^{2}\right)^{3 / 2}} $

${=-\frac{1}{3 x^{3}}(\sqrt{1-x^{2}})^{3}} $

$ \Rightarrow {(A(x))^3} = {\left( { - \frac{1}{{3{x^3}}}} \right)^3} = \frac{1}{{27{x^9}}}$

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