If ∫sec4/3⁡xcosec8/3⁡xdx=a(tan⁡x)−5/3+b(tan⁡x)1/3+C, then 5a+b=

If sec4/3xcosec8/3xdx=a(tanx)5/3+b(tanx)1/3+C, then 5a+b=

  1. A

    3

  2. B

    -3

  3. C

    0

  4. D

    -1

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    Solution:

    We have,

              Isec4/3xcosec8/3xdx I=cos4/3xsin8/3xdx=1cos4/3xsin8/3xdx

    The sum of the exponents of sin x and cos x is - 4, an even integer. So, we divide both numerator and denominator by cos4x.

     I    =sec4xtan8/3xdx=1+tan2x(tanx)8/3d(tanx) I    =(tanx)8/3+(tanx)2/3d(tanx) I    =35(tanx)5/3+3(tanx)1/3+C

     a=35 and b=35a+b=0

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