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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{{{x}^{2}}}{{{(9-{{x}^{2}})}^{3/2}}}\ dx=}\]

    A) \[\frac{x}{\sqrt{9-{{x}^{2}}}}-{{\sin }^{-1}}\frac{x}{3}+c\]

    B) \[\frac{x}{\sqrt{9-{{x}^{2}}}}+{{\sin }^{-1}}\frac{x}{3}+c\]

    C) \[{{\sin }^{-1}}\frac{x}{3}-\frac{x}{\sqrt{9-{{x}^{2}}}}+c\]

    D) None of these

    Correct Answer: A

    Solution :

    • Put \[x=3\sin \theta \Rightarrow dx=3\cos \theta \,d\theta ,\] therefore                   
    • \[\int_{{}}^{{}}{\frac{{{x}^{2}}}{{{(9-{{x}^{2}})}^{3/2}}}\,dx}=\int_{{}}^{{}}{\frac{9{{\sin }^{2}}\theta }{{{(9-9{{\sin }^{2}}\theta )}^{3/2}}.\,3\cos \theta }\,d\theta }\]
    • \[=\int_{{}}^{{}}{\frac{27{{\sin }^{2}}\theta \cos \theta }{27{{\cos }^{3}}\theta }}\,d\theta =\int_{{}}^{{}}{{{\tan }^{2}}\theta \,d\theta }=\int_{{}}^{{}}{({{\sec}^{2}}\theta -1)\,d\theta }\]                   
    • \[=\tan \theta -\theta +c=\tan \left\{ {{\sin }^{-1}}\left( \frac{x}{3} \right) \right\}-{{\sin }^{-1}}\left( \frac{x}{3} \right)+c\]                   
    • \[=\tan {{\tan }^{-1}}\left( \frac{\left( \frac{x}{3} \right)}{\sqrt{1-({{x}^{2}}/9)}} \right)-{{\sin }^{-1}}\left( \frac{x}{3} \right)+c\]                
    • \[=\frac{x}{\sqrt{9-{{x}^{2}}}}-{{\sin }^{-1}}\left( \frac{x}{3} \right)+c.\]


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