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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{x{{\sin }^{-1}}x\ dx}=\]  [MP PET 1991]

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

    B) \[\left( \frac{{{x}^{2}}}{2}+\frac{1}{4} \right){{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-{{x}^{2}}}+c\]

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

    D) \[\left( \frac{{{x}^{2}}}{2}+\frac{1}{4} \right){{\sin }^{-1}}x-\frac{x}{4}\sqrt{1-{{x}^{2}}}+c\]

    Correct Answer: A

    Solution :

    • \[\int_{{}}^{{}}{x{{\sin }^{-1}}xdx=\frac{{{x}^{2}}}{2}{{\sin }^{-1}}x-\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{x}^{2}}}}.\frac{{{x}^{2}}}{2}dx+c}}\]                    \[=\frac{{{x}^{2}}}{2}{{\sin }^{-1}}x-\frac{1}{2}\int_{{}}^{{}}{-\frac{(1-{{x}^{2}})+1}{\sqrt{1-{{x}^{2}}}}}dx+c\]                   
    • \[=\frac{{{x}^{2}}}{2}{{\sin }^{-1}}x+\frac{1}{2}\int_{{}}^{{}}{\sqrt{1-{{x}^{2}}}dx-\frac{1}{2}\int_{{}}^{{}}{\frac{1}{\sqrt{1-{{x}^{2}}}}dx+c}}\]                    \[=\frac{{{x}^{2}}}{2}{{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-{{x}^{2}}}+\frac{1}{4}{{\sin }^{-1}}x-\frac{1}{2}{{\sin }^{-1}}x+c\]                   
    • \[=\frac{{{x}^{2}}}{2}{{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-{{x}^{2}}}-\frac{1}{4}{{\sin }^{-1}}x\]                
    • \[=\left( \frac{{{x}^{2}}}{2}-\frac{1}{4} \right){{\sin }^{-1}}x+\frac{x}{4}\sqrt{1-{{x}^{2}}}+c\].


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