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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}}\ dx=\]

    A) \[\frac{1}{2}[{{\sin }^{-1}}{{x}^{2}}+\sqrt{1-{{x}^{4}}}]+c\] 

    B) \[\frac{1}{2}[{{\sin }^{-1}}{{x}^{2}}+\sqrt{1-{{x}^{2}}}]+c\]

    C) \[{{\sin }^{-1}}{{x}^{2}}+\sqrt{1-{{x}^{4}}}+c\]     

    D) \[{{\sin }^{-1}}{{x}^{2}}+\sqrt{1-{{x}^{2}}}+c\]

    Correct Answer: A

    Solution :

    • \[\int_{{}}^{{}}{x\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}}dx=\int_{{}}^{{}}{\frac{x.\,(1-{{x}^{2}})}{\sqrt{1-{{x}^{4}}}}}dx\] {Multiplying \[{{N}^{r}}\] and \[{{D}^{r}}\] by \[{{(1-{{x}^{2}})}^{1/2}}\}\]                   
    • \[=\int_{{}}^{{}}{\frac{x}{\sqrt{1-{{x}^{4}}}}}\,dx-\int_{{}}^{{}}{\frac{{{x}^{3}}}{\sqrt{1-{{x}^{4}}}}}dx\] \[=\frac{1}{2}[{{\sin }^{-1}}({{x}^{2}})+\sqrt{1-{{x}^{4}}}]+c\]. (By putting \[{{x}^{2}}=t\] and \[\sqrt{1-{{x}^{4}}}=\sqrt{t}\]respectively)


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