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

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.

Answer 1

$\displaystyle \int \cot^{n - 2} x\csc^2 x \,dx = -\displaystyle\int \cot^{n - 2} x \,d(\cot x) = -\dfrac{\cot^{n-1} x}{n - 1}$.

Or explicitly use the substitution $t = \cot x \Rightarrow dt = -\csc^2 x \cot x$.

Answer 2

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 $$ I_1=\int\cot x\ dx=\int\frac{\cos x}{\sin x}\ dx=\int\frac{1}{\sin x}\ d(\sin x)=\ln|\sin x|+C. $$

Answer 3

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}

Answer 4

Using your observation notice that:

$$I_{0}+I_{1}+2(I_{2}+...+I_{8})+I_{9}+I_{10}$$

$$=(I_{0}+I_{2})+(I_{1}+I_{3})+(I_{2}+I_{4})+...+(I_{7}+I_{9})+(I_{8}+I_{10})$$

$$=\sum_{n=0}^{8}(I_{n}+I_{n+2})=\sum_{n=0}^{8}\int\cot^{n}(x)\csc^{2}(x)dx=-\sum_{n=0}^{8}\int\cot^{n}(x)(-\csc^{2}(x))dx$$

$$=-\sum_{n=0}^{8}\int u^{n}du=-\sum_{n=0}^{8}\frac{\cot^{n+1}(x)}{n+1}$$