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

  • question_answer
    \[\int_{{}}^{{}}{\frac{\sin x\ dx}{{{(a+b\cos x)}^{2}}}=}\]

    A)            \[\frac{1}{b}(a+b\cos x)+c\]

    B)            \[\frac{1}{b(a+b\cos x)}+c\]

    C)            \[\frac{1}{b}\log (a+b\cos x)+c\]

    D)            None of these

    Correct Answer: B

    Solution :

                       Put \[a+b\cos x=t\Rightarrow dx=-\frac{dt}{b\sin x},\] then            \[\int_{{}}^{{}}{\frac{\sin x}{{{(a+b\cos x)}^{2}}}\,dx=-\frac{1}{b}\int_{{}}^{{}}{\frac{1}{{{t}^{2}}}\,dt}=\frac{1}{b}\frac{1}{t}+c}\]                                                                  \[=\frac{1}{b(a+b\cos x)}+c.\]


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