CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

  • question_answer
    \[\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{\sin x+\cos x}{1-{{\sin }^{2}}x} \right]}\,dx\]is

    A) \[({{e}^{x}}.\cos ecx)+C\]           

    B) \[{{e}^{x}}\cot x+C\]

    C) \[({{e}^{x}}.\sec x)+C\]

    D)  \[{{e}^{x}}\tan x+C\]

    Correct Answer: C

    Solution :

    \[I=\int{{{e}^{x}}\left( \frac{\sin x+\cos x}{1-{{\sin }^{2}}x} \right)dx}\] \[I=\int{{{e}^{x}}\left( \frac{\sin x+\cos x}{{{\cos }^{2}}x} \right)dx}\] \[I=\int{{{e}^{x}}\,\tan x.\sec x+dx+\int{\underset{II}{\mathop{{{e}^{x}}}}\,.\underset{I}{\mathop{\sec x}}\,\,\,dx}}\] \[I=\int{{{e}^{x}}\tan x.\sec x\,dx+(\sec \,x\,.\,{{e}^{x}}}\]                                 \[-\int{\sec x.\tan x\,{{e}^{x}}}dx\}\] \[I=\int{{{e}^{x}}\,\tan x.\sec x\,\,dx}+{{e}^{x}}\sec x\]                                 \[-\int{{{e}^{x}}\,\tan \,x.\,\sec x\,dx}\] \[I={{e}^{x}}.\,\sec \,\,x+c\]


You need to login to perform this action.
You will be redirected in 3 sec spinner