∫0nπ+w |sin⁡x|dx, where n∈N and 0≤w<π is equal to

0+w|sinx|dx, where nN and 0w<π is equal to

  1. A

    (2n + 1) + sin w

  2. B

    2n + cos w

  3. C

    (2n + 1) - cos w

  4. D

    None of these

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    Solution:

     I=0+w|sinx|dx  =0w|sinx|dx+w+w|sinx|dx=I1+I2I1=0w|sinx|dx=0wsinxdx =[cosx]0w=cosw+1=1coswI2=w+w|sinx|dx=n0π|(sinx)|dx =n0πsinxdx=n[cosx]0π=2n

    so, I=1cosw+2n=(2n+1)cosw

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