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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
    If\[f(x)=\tan x-{{\tan }^{3}}x+{{\tan }^{5}}x-.....\]to\[\infty \]with \[0<x<\frac{\pi }{4},\] then\[\int_{0}^{\frac{\pi }{4}}{f(x)}dx\]is equal to:

    A)  1            

    B)                         0

    C)  \[\frac{1}{4}\]                  

    D)         \[\frac{1}{2}\]

    E)  \[-\frac{1}{4}\]

    Correct Answer: C

    Solution :

    \[f(x)=\tan x-{{\tan }^{3}}x+{{\tan }^{5}}x-....\infty \] \[f(x)=\frac{\tan x}{1+{{\tan }^{2}}x}=\frac{\tan x}{{{\sec }^{2}}x}=\frac{\sin 2x}{2}\] \[\therefore \] \[\int_{0}^{\pi /4}{f(x)}dx=\int_{0}^{\pi /4}{\frac{\sin 2x}{2}}\]                 \[=\left[ -\frac{\cos 2x}{4} \right]_{0}^{\pi /4}\]                 \[=\frac{1}{4}\]


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