JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The integral \[\int_{{}}^{{}}{{{\sec }^{2/3}}}x\,\cos \,e{{c}^{4/3}}x\,dx\]is equal to (Hence C is a constant of integration) [JEE Main 9-4-2019 Morning]

    A) \[3{{\tan }^{-1/3}}x+C\]    

    B) \[-\frac{3}{4}{{\tan }^{-4/3}}x+C\]

    C) \[-3{{\cot }^{-1/3}}x+C\]   

    D) \[-3{{\tan }^{-1/3}}x+C\]

    Correct Answer: D

    Solution :

    \[I=\int_{{}}^{{}}{\frac{dx}{{{\left( \sin x \right)}^{4/3}}.{{\left( \cos x \right)}^{2/3}}}}\]           \[I=\int_{{}}^{{}}{\frac{dx}{{{\left( \frac{\sin x}{\cos x} \right)}^{4/3}}.{{\cos }^{2}}x}}\]           \[\Rightarrow I=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{{{\left( \tan x \right)}^{4/3}}}}dx\]           \[put\,\tan x=t\Rightarrow {{\sec }^{2}}xdx=dt\]           \[\therefore \]\[I=\int_{{}}^{{}}{\frac{dt}{{{t}^{4/3}}}\Rightarrow I=\frac{-3}{{{t}^{1/3}}}}+c\]           \[I=\frac{-3}{{{\left( \tan x \right)}^{1/3}}}+c\]

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