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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{4{{\sin }^{2}}x+5{{\cos }^{2}}x}=}\]              [AISSE 1986]

    A) \[\frac{1}{\sqrt{5}}{{\tan }^{-1}}\left( \frac{2\tan x}{\sqrt{5}} \right)+c\]

    B) \[\frac{1}{\sqrt{5}}{{\tan }^{-1}}\left( \frac{\tan x}{\sqrt{5}} \right)+c\]

    C) \[\frac{1}{2\sqrt{5}}{{\tan }^{-1}}\left( \frac{2\tan x}{\sqrt{5}} \right)+c\]

    D) None of these

    Correct Answer: C

    Solution :

    • \[\int_{{}}^{{}}{\frac{dx}{4{{\sin }^{2}}x+5{{\cos }^{2}}x}}=\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{4{{\tan }^{2}}x+5}=\frac{1}{4}\int_{{}}^{{}}{\frac{{{\sec }^{2}}x\,dx}{{{\tan }^{2}}x+\frac{5}{4}}}}\]                   
    • Put \[\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt,\] then it reduces to                   
    • \[\frac{1}{4}\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}+{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}=\frac{2}{4\sqrt{5}}{{\tan }^{-1}}\left( \frac{2t}{\sqrt{5}} \right)}+c\]
    • \[=\frac{1}{2\sqrt{5}}{{\tan }^{-1}}\left( \frac{2\tan x}{\sqrt{5}} \right)+c.\]


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