The value of ∫1sin3⁡xcos5⁡xdx, is 

The value of 1sin3xcos5xdx, is 

  1. A

    2tanx+23(tanx)3/2+C

  2. B

    2tanx23(tanx)3/2+C

  3. C

    2tanx+23(tanx)1/2+C

  4. D

    none of these

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    Solution:

    We have,

    I=1sin3xcos5xdx I=1sin3/2xcos52dx

     I=sec4xtan3/2xdx           [Dividing Nr and Dr by cos2 x]

     I=1+tan2xtan3/2xd(tanx) I=(tanx)3/2+(tanx)1/2d(tanx) I=2(tanx)1/2+23(tanx)3/2+C I=2tanx+23(tanx)3/2+C

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