TITLE: If $I_n =\int \cot^nx\ dx$ then $I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10}= $?
QUESTION [6 upvotes]: If $\displaystyle I_n =\int \cot^nx\ dx$ then find : 
$I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10} $ = ? 
My approach : 
$I_n = \displaystyle\int \cot^{n-2} \cot^2x dx$
$\Rightarrow I_n = \displaystyle\int \cot^{n-2} (\csc^2x -1)dx$ 
$\Rightarrow I_n = \displaystyle\int (\cot^{n-2} \csc^2x -\cot^{n-2} )dx$
$\Rightarrow I_n =\displaystyle \int( \cot^{n-2} \csc^2x) dx- I_{n-2}  $
$\Rightarrow I_n +I_{n-2} =\displaystyle \int( \cot^{n-2} \csc^2x) dx$
I am not getting how to integrate the RHS. now please guide thanks.

REPLY [2 votes]: Using your observation notice that:
$$I_{0}+I_{1}+2(I_{2}+...+I_{8})+I_{9}+I_{10}$$
$$=(I_{0}+I_{2})+(I_{1}+I_{3})+(I_{2}+I_{4})+...+(I_{7}+I_{9})+(I_{8}+I_{10})$$
$$=\sum_{n=0}^{8}(I_{n}+I_{n+2})=\sum_{n=0}^{8}\int\cot^{n}(x)\csc^{2}(x)dx=-\sum_{n=0}^{8}\int\cot^{n}(x)(-\csc^{2}(x))dx$$
$$=-\sum_{n=0}^{8}\int u^{n}du=-\sum_{n=0}^{8}\frac{\cot^{n+1}(x)}{n+1}$$