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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    The value of \[\int_{{}}^{{}}{\frac{\sin x}{{{\cos }^{2}}x}\ dx}\] is

    A)            \[\sin x+k\]

    B)            \[\tan x+k\]

    C)            \[\sec x+k\]

    D)            \[\tan x+\sec x+k\]

    Correct Answer: C

    Solution :

               Given \[I=\int_{{}}^{{}}{\frac{\sin x}{{{\cos }^{2}}x}}\,dx\]. Put \[\cos x=t\Rightarrow \sin x\,dx=-dt\]            \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{-dt}{{{t}^{2}}}}=\frac{1}{t}+k=\sec x+k\].


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