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If $I_{n} = \int \cot^{n} x d x$ , and $I_{0} + I_{1} + 2 (I_{2} + \dots + I_{8})$
$+ I_{9} + I_{10} = A (u + \frac{u^{2}}{2} + \dots + \frac{u^{9}}{9})$ , where $u = \cot x$ then

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A=1

A =– 1

A =1/2

A =– 1/2

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detailed solution

Correct option is B

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

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If $I_{n} = \int \cot^{n} x d x$ , and $I_{0} + I_{1} + 2 (I_{2} + \dots + I_{8})$
$+ I_{9} + I_{10} = A (u + \frac{u^{2}}{2} + \dots + \frac{u^{9}}{9})$ , where $u = \cot x$ then