If ∫3sin⁡x+2cos⁡x3cos⁡x+2sin⁡xdx=ax+blog⁡|3cos⁡x+2sin⁡x|+C, then 

If 3sinx+2cosx3cosx+2sinxdx=ax+blog|3cosx+2sinx|+C, then 

  1. A

    a=513, b=-1213

  2. B

    a=1213, b=-513

  3. C

    a=1213, b=513

  4. D

    a=-125, b=-513

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    Solution:

    Let 

        I=3sinx+2cosx3cosx+2sinxdx

    Let 3 sin x+2 cos x

          =λddx(3cosx+2sinx)+μ(3cosx+2sinx)

      3sinx+2cosx   =λ(3sinx+2cosx)+μ(3cosx+2sinx)

    Comparing the coefficients of sin x and cos x on both sides, we get

          3λ+2μ=3 and 2λ+3μ=2

     λ=5/13  and μ=12/13

     I    =513(3sinx+2cosx)+1213(3cosx+2sinx) I    =12131dx5133sinx+2cosx3cosx+2sinxdx I    =12131dx51313cosx+2sinxd(3cosx+2sinx) I    =1213x513log|3cosx+2sinx|+C

    Hence, a=1213, b=-513

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