The value of  2∫sin⁡xsin⁡x−π4dx, is

The value of  2sinxsinxπ4dx, is

  1. A

    x+logsinxπ4+C

  2. B

    xlogcosxπ4+C

  3. C

    x+logcosxπ4+C

  4. D

    xlogsinxπ4+C

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    Solution:

    We have,

    2sinxsinxπ4dx=2sinxπ4+π4sinxπ4dx=2sinxπ4cosπ4+cosxπ4sinπ4sinxπ4dx=1dx+cotxπ4dx=x+logsinxπ4+C

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