JEE Main & Advanced Sample Paper JEE Main - Mock Test - 6

  • question_answer
    The value of \[\int\limits_{0}^{\pi /2}{\frac{dx}{1+\cot x}}\] is

    A) \[\pi /4\]

    B) \[\pi /2\]

    C) 0

    D) \[\pi \]

    Correct Answer: A

    Solution :

    [a] : Let \[I=\int\limits_{0}^{\pi /2}{\frac{dx}{1+\cot x}}=\int\limits_{0}^{\pi /2}{\frac{\cos xdx}{\cos x+\sin x}}\]
    Replacing x by \[\frac{\pi }{2}-x\], we get
    \[I=\int\limits_{0}^{\pi /2}{\frac{\cos \left( \frac{\pi }{2}-x \right)dx}{\cos \left( \frac{\pi }{2}-x \right)+\sin \left( \frac{\pi }{2}-x \right)}}\]
    \[I=\int\limits_{0}^{\pi /2}{\frac{\sin x\,dx}{\sin x+\cos x}}\]
    \[\therefore \]\[I+I=\int\limits_{0}^{\pi /2}{\frac{\cos xdx}{\cos x+\sin x}}+\int\limits_{0}^{\pi /2}{\frac{\sin xdx}{\sin x+\cos x}}\]
    \[=\int\limits_{0}^{\pi /2}{dx}=[x]_{0}^{\pi /2}=\frac{\pi }{2}\]
    So,\[2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\Rightarrow \int\limits_{0}^{\pi /2}{\frac{dx}{1+\cot x}=\frac{\pi }{4}}\]

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