JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int_{{}}^{{}}{\sqrt{\frac{x}{{{a}^{3}}-{{x}^{3}}}}\ dx=}\]

    A)            \[{{\sin }^{-1}}{{\left( \frac{x}{a} \right)}^{3/2}}+c\]

    B)            \[\frac{2}{3}{{\sin }^{-1}}{{\left( \frac{x}{a} \right)}^{3/2}}+c\]

    C)            \[\frac{3}{2}{{\sin }^{-1}}{{\left( \frac{x}{a} \right)}^{3/2}}+c\]

    D)            \[\frac{3}{2}{{\sin }^{-1}}{{\left( \frac{x}{a} \right)}^{2/3}}+c\]

    Correct Answer: B

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner