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Q.

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

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a

A=1

b

A =– 1

c

A =1/2

d

A =– 1/2

answer is B.

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Detailed Solution

In+In+2=∫cotn⁡xcosec2⁡xdx=−1n+1cotn+1⁡xNow , I0+I1+2I2+I3+⋯+I8+I9+I10=I0+I2+I1+I3+I2+I4+⋯+I7+I9+I8+I10=−u+12u2+13u3+⋯+19u9where , u=cot⁡xThus , A=-1
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