A) \[\frac{1}{2}{{\sin }^{-1}}({{x}^{4}})+c\]
B) \[\frac{1}{3}{{\sin }^{-1}}({{x}^{4}})+c\]
C) \[\frac{1}{4}{{\sin }^{-1}}({{x}^{4}})+c\]
D) None of these
Correct Answer: C
Solution :
\[\int_{{}}^{{}}{\frac{{{x}^{3}}}{\sqrt{1-{{x}^{8}}}}\,dx=\int_{{}}^{{}}{\frac{{{x}^{3}}}{\sqrt{1-{{({{x}^{4}})}^{2}}}}}}\] Put \[{{x}^{4}}=t\Rightarrow 4{{x}^{3}}dx=dt,\] then it reduces to \[\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{\sqrt{1-{{x}^{2}}}}=\frac{1}{4}[{{\sin }^{-1}}(t)]+c=\frac{1}{4}{{\sin }^{-1}}({{x}^{4}})+c.}\]
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