CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    \[\int_{0}^{\frac{\pi }{2}}{\frac{\sin x}{\sin x+\cos x}}dx\]equals to:

    A)  \[\pi \]

    B)                                                         \[\frac{\pi }{2}\]

    C)         \[\frac{\pi }{3}\]                                             

    D)         \[\frac{\pi }{4}\]

    E)         \[\frac{\pi }{5}\]

    Correct Answer: D

    Solution :

    Let\[I=\int_{0}^{\pi /2}{\frac{\sin x}{\sin x+\cos x}dx}\]      ...(i) \[I=\int_{0}^{\pi /2}{\frac{\sin \left( \frac{\pi }{2}-x \right)}{\sin \left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}dx}\] \[I=\int_{0}^{\pi /2}{\frac{\cos x}{\sin x+\cos x}dx}\]                      ?. (ii) On adding Eqs. (i) and (ii), we get                 \[2I=\int_{0}^{\pi /2}{1}\,dx=\frac{\pi }{2}\] \[\Rightarrow \]               \[I=\frac{\pi }{4}\]


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