JEE Main & Advanced Sample Paper JEE Main Sample Paper-8

  • question_answer
    Find \[\int{{{e}^{\sin \,x}}\left( \frac{x{{\cos }^{3}}x-\sin x}{{{\cos }^{2}}x} \right)}dx\]

    A)  \[x{{e}^{\sin x}}-{{e}^{\sin \,x}}\sec x+C\]

    B)  \[x{{e}^{\cos x}}-{{e}^{\sin \,x}}\sec x+C\]

    C)   \[{{x}^{2}}{{e}^{\sin x}}+{{e}^{\sin \,x}}\sec x+C\]

    D)  \[2{{x}^{2}}{{e}^{\sin x}}-{{e}^{\sin \,x}}\tan x+C\]

    Correct Answer: A

    Solution :

     \[\int_{{}}^{{}}{{{e}^{\sin x}}}\left( \frac{x{{\cos }^{3}}x-\sin x}{{{\cos }^{2}}x} \right)dx\] \[=\int_{{}}^{{}}{{{e}^{\sin x}}}x\cos xdx-\int_{{}}^{{}}{{{e}^{\sin x}}}\tan x\sec xdx\] \[=\int_{{}}^{{}}{xd\left( {{e}^{\sin x}} \right)}-\int_{{}}^{{}}{{{e}^{\sin x}}}d(\sec x)\] \[=\left\{ x{{e}^{\sin x}}-\int_{{}}^{{}}{{{e}^{\sin x}}}dx \right\}\] \[=\left\{ {{e}^{\sin x}}\sec x-\int_{{}}^{{}}{{{e}^{\sin x}}}\sec x\cos xdx \right\}\] \[=x{{e}^{\sin x}}-{{e}^{\sin x}}\sec x+C\]


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