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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\sqrt{\frac{a-x}{x}}\ dx=}\]

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

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

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

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

    Correct Answer: A

    Solution :

    • \[I=\int_{{}}^{{}}{\sqrt{\frac{a-x}{x}}\,dx}\].                   
    • Put \[x=a{{\sin }^{2}}\theta \Rightarrow dx=2a\sin \theta \cos \theta \,d\theta ,\] then                   
    • \[I=\int_{{}}^{{}}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}}\,.\,2a\sin \theta \cos \theta \,d\theta \]                     
    • \[=a\int_{{}}^{{}}{2{{\cos }^{2}}\theta \,d\theta }=a\int_{{}}^{{}}{(1+\cos 2\theta )\,d\theta }\]                  
    • \[=a\,\left[ {{\sin }^{-1}}\sqrt{\frac{x}{a}}+\sqrt{\frac{x}{a}}\,.\,\sqrt{\frac{a-x}{a}} \right]+c\].


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