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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}=}\]                

    A) \[\text{cosec}\,\,(a-b)\log \frac{\sin (x-a)}{\sin (x-b)}+c\]

    B) \[\text{cosec}(a-b)\log \frac{\cos (x-a)}{\cos (x-b)}+c\]

    C) \[\text{cosec}(a-b)\log \frac{\sin (x-b)}{\sin (x-a)}+c\]       

    D) \[\text{cosec}(a-b)\log \frac{\cos (x-b)}{\cos (x-a)}+c\]

    Correct Answer: B

    Solution :

    • \[\int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}}\]                                
    • \[=\frac{1}{\sin (a-b)}\int_{{}}^{{}}{\frac{\sin \left\{ (x-b)-(x-a) \right\}}{\cos (x-a)\,.\,\cos (x-b)}\,dx}\]                                
    • \[=\frac{1}{\sin (a-b)}\int_{{}}^{{}}{\left\{ \frac{\sin (x-b)}{\cos (x-b)}-\frac{\sin (x-a)}{\cos (x-a)} \right\}dx}\]                                
    • \[=\text{cosec}\,(a-b)\log \frac{\cos (x-a)}{\cos (x-b)}+c\].


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