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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    \[{{I}_{n}}=\int_{\,0}^{\,\pi /4}{{{\tan }^{n}}x\,dx}\], then \[\underset{n-\infty }{\mathop{\lim }}\,n\,[{{I}_{n}}+{{I}_{n-2}}]\] equals [AIEEE 2002]

    A) 1/2

    B) 1

    C) \[\infty \]                              

    D) 0

    Correct Answer: B

    Solution :

    • \[{{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}\,dx=\int_{0}^{\pi /4}{\,({{\sec }^{2}}x-1){{\tan }^{n-2}}x\,dx}\]                       
    • \[=\int_{0}^{\pi /4}{{{\sec }^{2}}x{{\tan }^{n-2}}}x\,dx-\int_{0}^{\pi /4}{{{\tan }^{n-2}}}x\,dx\]                      
    • \[=\left[ \frac{{{\tan }^{n-1}}x}{n-1} \right]_{0}^{\pi /4}-{{I}_{n-2}}\] Þ \[{{I}_{n}}+{{I}_{n-2}}=\frac{1}{n-\,1}\]                   
    • Now, \[\underset{n\to \infty }{\mathop{\lim }}\,\,n[{{I}_{n}}+{{I}_{n-2}}]\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{n}{n-1}\]=\[\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{1}{1-\frac{1}{n}}=1\].


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