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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Critical Thinking

  • question_answer
    \[\int{\frac{dx}{\sin x-\cos x+\sqrt{2}}}\] equals  [MP PET 2002]

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

    B) \[\frac{1}{\sqrt{2}}\tan \left( \frac{x}{2}+\frac{\pi }{8} \right)+c\]

    C) \[\frac{1}{\sqrt{2}}\cot \left( \frac{x}{2}+\frac{\pi }{8} \right)+c\]

    D) \[-\frac{1}{\sqrt{2}}\cot \left( \frac{x}{2}+\frac{\pi }{8} \right)+c\]

    Correct Answer: D

    Solution :

    • \[I=\int{\frac{dx}{\sin x-\cos x+\sqrt{2}}}\]             
    • \[=\int{\frac{dx}{\sqrt{2}(\sin x.\sin \frac{\pi }{4}-\cos x\cos \frac{\pi }{4}+1)}}\]              
    • \[=\frac{1}{\sqrt{2}}\int{\frac{dx}{1-\cos (x+\frac{\pi }{4})}}\]\[=\frac{1}{\sqrt{2}}\int{\frac{dx}{1-\cos 2\left( \frac{x}{2}+\frac{\pi }{8} \right)}}\]  
    • \[=\frac{1}{\sqrt{2}}\int{\frac{dx}{2{{\sin }^{2}}\left( \frac{x}{2}+\frac{\pi }{8} \right)}}\] 
    • \[=\frac{1}{2\sqrt{2}}\int{\text{cose}{{\text{c}}^{2}}\left( \frac{x}{2}+\frac{\pi }{8} \right)\,dx}\]                  
    • \[=\frac{1}{2\sqrt{2}}\frac{-\cot \,\left( \frac{x}{2}+\frac{\pi }{8} \right)\,}{1/2}+c\]\[=\frac{-1}{\sqrt{2}}\cot \,\left( \frac{x}{2}+\frac{\pi }{8} \right)\,+c\].


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