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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{{{\tan }^{3}}}2x\sec 2x\ dx=\]      [IIT 1977]

    A) \[\frac{1}{6}{{\sec }^{3}}2x-\frac{1}{2}\sec 2x+c\]

    B) \[\frac{1}{6}{{\sec }^{3}}2x+\frac{1}{2}\sec 2x+c\]

    C) \[\frac{1}{9}{{\sec }^{2}}2x-\frac{1}{3}\sec 2x+c\]

    D) None of these

    Correct Answer: A

    Solution :

    • \[\int_{{}}^{{}}{{{\tan }^{3}}2x\sec 2x\,dx}=\int_{{}}^{{}}{({{\sec }^{2}}2x-1)\sec 2x\tan 2x\,dx}\]                   
    • \[\int_{{}}^{{}}{({{\sec }^{3}}2x\tan 2x-\sec 2x\tan 2x)dx}\]                   
    • \[=\int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x\,dx}-\int_{{}}^{{}}{\sec 2x\tan 2x\,dx}\]     ..?(i)                   
    • Now, we take \[\int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x\,dx}\]                   
    • Put \[\sec 2x=t\Rightarrow \sec 2x\tan 2x=\frac{dt}{2},\] then it reduces to                   
    • \[\frac{1}{2}\int_{{}}^{{}}{{{t}^{2}}dt}=\frac{{{t}^{3}}}{6}=\frac{{{\sec }^{3}}2x}{6}\]                   
    • From (i), \[\int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x\,dx}-\int_{{}}^{{}}{\sec 2x\tan 2x\,dx}\]                   
    • \[=\frac{{{\sec }^{3}}2x}{6}-\frac{\sec 2x}{2}+c.\]                   
    • Trick : Let \[\sec 2x=t,\] then \[\sec 2x\tan 2x\,dx=\frac{1}{2}dt\]                
    • \\[\frac{1}{2}\int_{{}}^{{}}{({{t}^{2}}-1)\,dt}=\frac{1}{6}{{t}^{3}}-\frac{1}{2}t+c=\frac{1}{6}{{\sec }^{3}}2x-\frac{1}{2}\sec 2x+c\].


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