A) \[\frac{1}{9{{x}^{4}}}\]
B) \[\frac{-1}{3{{x}^{3}}}\]
C) \[\frac{1}{27{{x}^{6}}}\]
D) \[\frac{-1}{27{{x}^{9}}}\]
Correct Answer: D
Solution :
Let\[I=\int_{{}}^{{}}{\frac{\sqrt{1-{{x}^{2}}}}{{{x}^{4}}}}dx=\int_{{}}^{{}}{\frac{\sqrt[x]{\frac{1}{{{x}^{2}}}-1}}{{{x}^{4}}}}dx\] Put\[\frac{1}{{{x}^{2}}}-1=t\Rightarrow \frac{-2}{{{x}^{3}}}dx=dt\Rightarrow \frac{dx}{{{x}^{3}}}=-\frac{1}{2}dt\] \[\therefore \]\[I=-\frac{1}{2}\int_{{}}^{{}}{\sqrt{t}}dt=-\frac{1}{3}({{t}^{3/2}})+c\] \[=-\frac{1}{3}{{\left( \frac{1}{{{x}^{2}}}-1 \right)}^{3/2}}+c=-\frac{1}{3{{x}^{3}}}{{\left( 1-{{x}^{2}} \right)}^{3/2}}+c\] \[={{\frac{\left( \sqrt{1-{{x}^{2}}} \right)}{-3{{x}^{3}}}}^{3}}+c\] \[\therefore \]\[A(x)=-\frac{1}{3{{x}^{3}}}\]and\[m=3\] \[\therefore \]\[{{(A(x))}^{3}}={{\left( \frac{-1}{3{{x}^{3}}} \right)}^{3}}=-\frac{1}{27{{x}^{9}}}\]
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