If In=∫cotn⁡xdx , and  I0+I1+2I2+…+I8+I9+I10=Au+u22+…+u99 , where u=cot⁡x then 

If In=cotnxdx , and  I0+I1+2I2++I8

+I9+I10=Au+u22++u99 , where u=cotx then 

  1. A

    A=1

  2. B

    A = 1

  3. C

    A =1/2 

  4. D

    A = 1/2

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    Solution:

     In+In+2=cotnxcosec2xdx=1n+1cotn+1x

    Now , 

    I0+I1+2I2+I3++I8+I9+I10=I0+I2+I1+I3+I2+I4++I7+I9+I8+I10=u+12u2+13u3++19u9

    where , u=cotx

    Thus , A=-1

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