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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{2+\cos x}=}\]

    A) \[2{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)+c\]

    B) \[\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)+c\]

    C) \[\frac{1}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{1}{\sqrt{3}}\tan \frac{x}{2} \right)+c\]

    D) None of these

    Correct Answer: B

    Solution :

    • \[\int_{{}}^{{}}{\frac{dx}{2+\cos x}=\int_{{}}^{{}}{\frac{dx}{2{{\sin }^{2}}\left( \frac{x}{2} \right)+2{{\cos }^{2}}\left( \frac{x}{2} \right)+{{\cos }^{2}}\left( \frac{x}{2} \right)-{{\sin }^{2}}\left( \frac{x}{2} \right)}}}\]                   
    • \[=\int_{{}}^{{}}{\frac{dx}{{{\sin }^{2}}\left( \frac{x}{2} \right)+3{{\cos }^{2}}\left( \frac{x}{2} \right)}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}\left( \frac{x}{2} \right)}{{{\tan }^{2}}\left( \frac{x}{2} \right)+3}dx}\]                   
    • Put \[\tan \left( \frac{x}{2} \right)=t\Rightarrow {{\sec }^{2}}\left( \frac{x}{2} \right)\,dx=2dt,\] then it reduces to                
    • \[2\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+3}}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{t}{\sqrt{3}} \right)+c=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{\tan \left( \frac{x}{2} \right)}{\sqrt{3}} \right)+c\].


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