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

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

1. A

$A=1$

2. B

3. C

4. D

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### Solution:

${I}_{n}+{I}_{n+2}=\int {\mathrm{cot}}^{n}x{\mathrm{cosec}}^{2}xdx=-\frac{1}{n+1}{\mathrm{cot}}^{n+1}x$

Now ,

$\begin{array}{l}{I}_{0}+{I}_{1}+2\left({I}_{2}+{I}_{3}+\cdots +{I}_{8}\right)+{I}_{9}+{I}_{10}\\ =\left({I}_{0}+{I}_{2}\right)+\left({I}_{1}+{I}_{3}\right)+\left({I}_{2}+{I}_{4}\right)\\ +\cdots +\left({I}_{7}+{I}_{9}\right)+\left({I}_{8}+{I}_{10}\right)\\ =-\left(u+\frac{1}{2}{u}^{2}+\frac{1}{3}{u}^{3}+\cdots +\frac{1}{9}{u}^{9}\right)\end{array}$

where , $u=\mathrm{cot}x$

Thus , $A=-1$

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