CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[\int{\frac{(\sin \theta +\cos \theta )}{\sqrt{\sin 2\theta }}}d\theta \]is equal to:

    A)  \[\log |\cos \theta -\sin \theta +\sqrt{\sin 2\theta }|+c\]

    B)  \[\log |\sin \theta -\cos \theta +\sqrt{\sin 2\theta }|+c\]

    C)  \[{{\sin }^{-1}}(\sin \theta -\cos \theta )+c\]

    D)  \[{{\sin }^{-1}}(\sin \theta +\cos \theta )+c\]

    E)  \[{{\sin }^{-1}}(\cos \theta -\sin \theta )+c\]

    Correct Answer: C

    Solution :

    Let\[I=\int{\frac{\sin \theta +\cos \theta }{\sqrt{1+\sin 2\theta -1}}d\theta }\] \[=\int{\frac{\sin \theta +\cos \theta }{\sqrt{1-{{(\sin \theta -\cos \theta )}^{2}}}}d\theta }\] Let         \[\sin \theta -\cos \theta =t\] \[\Rightarrow \] \[(\cos \theta +\sin \theta )d\theta =dt\] \[\therefore \] \[I=\int{\frac{1}{\sqrt{1-{{t}^{2}}}}}dt={{\sin }^{-1}}(\sin \theta -\cos \theta )+c\]


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