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.
 A: Hint :
The strategy for integrating by reduction formula of $\cot^n x$ for integer $n \ge 2$ is to start by factoring out one $\cot^2 x$ term.
\begin{align}
\int\cot^n x\ dx&=\int\cot^{n-2} x\cot^2x\ dx\\
&=\int\cot^{n-2} x(\csc^2x-1)\ dx\\
&=\int\cot^{n-2} x\csc^2x\ dx-\int\cot^{n-2} x\ dx.
\end{align}
The derivative of  $\cot x$  is  $-\csc^2 x$,  so in the first integral, the derivative is present and the integral is easily evaluated using the power rule. Hence
$$
 I_n=\int\cot^n x\ dx=\color{blue}{-\frac{1}{n-1}\cot^{n-1} x-\int\cot^{n-2} x\ dx},
$$
and
$$
I_1=\int\cot x\ dx=\int\frac{\cos x}{\sin x}\ dx=\int\frac{1}{\sin x}\ d(\sin x)=\ln|\sin x|+C.
$$
A: 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}
A: Using your observation notice that:
$$I_{0}+I_{1}+2(I_{2}+...+I_{8})+I_{9}+I_{10}$$
$$=(I_{0}+I_{2})+(I_{1}+I_{3})+(I_{2}+I_{4})+...+(I_{7}+I_{9})+(I_{8}+I_{10})$$
$$=\sum_{n=0}^{8}(I_{n}+I_{n+2})=\sum_{n=0}^{8}\int\cot^{n}(x)\csc^{2}(x)dx=-\sum_{n=0}^{8}\int\cot^{n}(x)(-\csc^{2}(x))dx$$
$$=-\sum_{n=0}^{8}\int u^{n}du=-\sum_{n=0}^{8}\frac{\cot^{n+1}(x)}{n+1}$$
A: $\displaystyle \int \cot^{n - 2} x\csc^2 x \,dx = -\displaystyle\int \cot^{n - 2} x \,d(\cot x) = -\dfrac{\cot^{n-1} x}{n - 1}$.
Or explicitly use the substitution $t = \cot x \Rightarrow dt = -\csc^2 x \cot x$.
