The value of ∫03π2 tan−1⁡tan⁡x−sin−1⁡sin⁡xtan−1⁡tan⁡x+sin−1⁡sin⁡xdx is equal to

The value of 03π2tan1tanxsin1sinxtan1tanx+sin1sinxdx is equal to

  1. A

    π2

  2. B

    π

  3. C

    3π2

  4. D

    None of these

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    Solution:

    0π/2tan1tanxsin1sin xtan1tanx+sin1sinxdx+π/23π/2tan1tanxsin1sin xtan1tanx+sin1sin1xdx 

    Integrand, is discontinuous π2, then

    0π/20dx+π/23π/20dx=00<x<π2,tan1tanx=sin1sinxand π2<x<3π2,tan1tanx=sin1sinx

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