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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    Let \[n\ge 2\] be a natural number and \[0<\theta <\frac{\pi }{2}\]Then \[\int{\frac{(si{{n}^{n}}\,\theta \,-\,sin\,\theta ){{\,}^{1/n}}\,\cos \,\theta }{\sin {{\,}^{n+1}}\,\theta }}\,d\theta \] is equal to -(where C is a constant of integration) [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1+\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]

    B) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n+1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]

    C) \[\frac{n}{{{n}^{2}}-1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]

    D) \[\frac{n}{{{n}^{2}}+1}\,{{\left( 1-\frac{1}{\sin {{\,}^{n-1}}\,\theta } \right)}^{\frac{n+1}{n}}}\,+C\]

    Correct Answer: C

    Solution :

    put sin \[\theta \] \[=t\Rightarrow \,\,cos\,\theta d\,\theta =dt\] \[I=\int{\frac{{{({{t}^{n}}-t)}^{1/n}}}{{{t}^{n+1}}}dt}\,\,=\,\,\int{\frac{{{\left( 1-\frac{1}{{{t}^{n-1}}} \right)}^{1/n}}\,dt}{{{t}^{n}}}}\] \[=\,\,\int{{{x}^{1/n}}\frac{dx}{(n-1)}\,\,\,\,\,\,\,\,\,\,\,\,\,put\,1-\frac{1}{{{t}^{n-1}}}\,=x}\] \[=\frac{1}{(n-1)}\frac{{{x}^{\frac{1}{n}+1}}}{\frac{1}{n+1}}+C\,\,\,\Rightarrow \,\,(n-1){{t}^{-(n-1)-1}}\,dt\,=\,dx\] \[\Rightarrow \,\,\,(n-1)\frac{dt}{{{t}^{n}}}\,\,=\,\,dx\] \[=\,\,\,\,\,\,\frac{n}{{{n}^{2}}-1}{{\left( 1-\frac{1}{{{\sin }^{n-1}}\theta } \right)}^{\frac{n+1}{n}}}+C\]


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