JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Mock Test - Integrals

  • question_answer
    If\[\int \frac{3\sin x+2\cos x}{3\cos x+2\sin x}dx=ax+b\] ln\[\left| 2\sin x+3\cos x \right|+C\], then

    A) \[a=-\frac{12}{13},b=\frac{15}{39}\]

    B) \[a=-\frac{7}{13},b=\frac{6}{13}\]

    C) \[a=\frac{12}{13},b=-\frac{15}{39}\]

    D) \[a=-\frac{7}{13},b=-\frac{6}{13}\]

    Correct Answer: C

    Solution :

    [c] Differentiating both sides, we get \[\frac{3\sin x+2\cos x}{3\cos x+2\sin x}=\]\[a+\frac{b(2\cos x-3\sin x)}{(2\sin x+3\cos x)}\] \[=\frac{\sin x(2a-3b)+\cos x\,(3a+2b)}{(3\cos x+2\sin x)}\] Comparing like terms on both sides, we get \[3=2a-3b,2=3a+2b\] Or \[a=\frac{12}{13},b=-\frac{15}{39}\]


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