JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    \[\int{\frac{dx}{\cos x-\sin x}}\]is equal to

    A) \[\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{x}{2}-\frac{\pi }{8} \right) \right|+C\]

    B) \[\frac{1}{\sqrt{2}}\log \left| \cot \left( \frac{x}{2} \right) \right|+C\]

    C) \[\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{x}{2}-\frac{3\pi }{8} \right) \right|+C\]

    D) \[\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{x}{2}+\frac{3\pi }{8} \right) \right|+C\]

    Correct Answer: D

    Solution :

    Let\[l=\int{\frac{dx}{\cos x-\sin x}}\] \[=\frac{1}{\sqrt{2}}\int{\frac{dx}{\left( \frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x \right)}}\] \[=\frac{1}{\sqrt{2}}\int{\frac{dx}{\cos \left( x+\frac{\pi }{4} \right)}}=\frac{1}{\sqrt{2}}\int{\sec \left( x+\frac{\pi }{4} \right)}dx\] \[=\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{\pi }{4}+\frac{x}{2}+\frac{\pi }{8} \right) \right|+C\] \[=\frac{1}{\sqrt{2}}\log \left| \tan \left( \frac{x}{2}+\frac{3\pi }{8} \right) \right|+C\]


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