<s>[INST] If $I_n =\int \cot^nx\ dx$ then $I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10}= $? 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.
 [/INST] Consider using the integral 
\begin{align}
I_{n} = \int \cot^{n}(x) \ dx
\end{align}
to evaluate the integral $J = I_{0} + I_{1} + 2(I_{3} + \cdots + I_{8}) + I_{9} + I_{10}$. Let $P$ be the integrand of $J$ for which
\begin{align}
P &= 1 + \cot^{1}(x) + 2( \cot^{2}(x) + \cdots + \cot^{8}(x)) + \cot^{9}(x) + \cot^{10}(x) \\
&= \sum_{k=0}^{10} \cot^{k}(x) + \cot^{2}(x) \ \sum_{k=0}^{6} \cot^{k}(x) \\
&= (1 + \cot^{2}(x)) \ \frac{1-\cot^{9}(x)}{1-\cot(x)} \\
&= \frac{1-\cot^{9}(x)}{(1-\cot(x)) \ \sin^{2}(x)}.
\end{align}
Now,
\begin{align}
J &= \int \frac{1-\cot^{9}(x)}{(1-\cot(x)) \ \sin^{2}(x)} \ dx \\
&= - \int \frac{1 - u^{9}}{1-u} \ du \\
&= - \sum_{k=1}^{9} \frac{u^{k}}{k} \\
J &= - \sum_{k=1}^{9} \frac{\cot^{k}(x)}{k}.
\end{align}
</s>